12th NCERT Continuity and Differentiability Exercise 5.5 Questions 18
Differentiate the functions given in Exercises 1 to 11 w.r.t x

Question (1)

cosx. cos2x.cos3x

Solution

y = cosx. cos2x.cos3x
$\log abc = \log a + \log b + \log c$ $\frac{d}{{dx}}\log x = \frac{1}{x}$
Take log on both sides
logy= log[cosx. cos2x.cos3x]
logy = log cosx + log cos 2x + log cos 3x
Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{\cos x}}\frac{d}{{dx}}\cos x + \frac{1}{{\cos 2x}}\frac{d}{{dx}}\cos 2x + \frac{1}{{\cos 3x}}\frac{d}{{dx}}\cos 3x$ $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{\cos x}}\left( { - \sin x} \right) + \frac{1}{{\cos 2x}}\left( { - 2\sin 2x} \right) + \frac{1}{{\cos 3x}}\left( { - 3\sin 3x} \right)$ $\frac{{dy}}{{dx}} = -y\left[ {\tan x + 2\tan 2x + 3\tan 3x} \right]$ $\frac{{dy}}{{dx}} = - \cos x\cos 2x\cos 3x\left[ {\tan x + 2\tan 2x + 3\tan 3x} \right]$

Question (2)

$\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}}$

Solution

$\log \left( {ab} \right) = \log a + \log b$ $\log \left( {\frac{a}{b}} \right) = \log a - \log b$ $\frac{d}{{dx}}\log x = \frac{1}{x}$
Take log on both sides $\log y = \frac{1}{2}\left[ {\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)} \right]$ Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$ $\frac{{dy}}{{dx}} = \frac{y}{2}\left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$ $\frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {\frac{{(x - 1)(x - 2)}}{{(x - 3)(x - 4)(x - 5)}}} \left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$

Question (3)

(log x)cosx

Solution

$\log {a^m} = m\log a$ $\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$
Take log on both sides
log y = log [(log x)cos x]
log y = cosx log(logx)
Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \cos x\frac{d}{{dx}}\log \left( {\log x} \right) + \log \left( {\log x} \right)\frac{d}{{dx}}\cos x$ $\frac{1}{y}\frac{{dy}}{{dx}} = \cos x\frac{1}{{\log x}} \times \frac{1}{x} + \log \left( {\log x} \right)\left( { - \sin x} \right)$ $\frac{{dy}}{{dx}} = y\left[ {\frac{{\cos x}}{{x\log x}} + - \sin x\log \left( {\log x} \right)} \right]$ $\frac{{dy}}{{dx}} = {\left( {\log x} \right)^{\cos x}}\left[ {\frac{{\cos x}}{{x\log x}} - \sin x\log \left( {\log x} \right)} \right]$

Question (4)

xx -2sinx

Solution

Here variable raised to variable is given so we have to take log on both sides, but log(a+b) does not have any property. So we assume each one as new variable and then solve separately $\log {a^m} = m\log a$ $\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$ $\frac{d}{{dx}}{a^x} = {a^x}\log a$
let xx = u
Take log on both sides
logu= logxx
log u = xlogx
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x$ $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{1}{x} + \log x \cdot \left( 1 \right)$ $\frac{1}{u}\frac{{du}}{{dx}} = 1 + \log x$ $\frac{{du}}{{dx}} = u\left( {1 + \log x} \right) = {x^x}\left( {1 + \log x} \right)$ y = xx - 2sinx
y = u - 2sinx
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} - \frac{d}{{dx}}{2^{\sin x}}$ $\frac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right) - {2^{\sin x}}\log _e^2\frac{d}{{dx}}\sin x$ $\frac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right) - {2^{\sin x}}\left( {\cos x} \right)\log _e^2$

Question (5)

(x+3)2 . (x+4)3 . (x+5)4

Solution

Here polynomial with three terms with different power is given. So we take log on both side.
log ab = log a + log b
logam = mloga
logy = log[(x+3)2 . (x+4)3 . (x+5)4]
log y = 2log(x+3) + 3log(x+4) + 4log(x+5) Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \left[ {\frac{2}{{x + 3}} + \frac{3}{{x + 4}} + \frac{4}{{x + 5}}} \right]$ $\frac{{dy}}{{dx}} = y\left[ {\frac{{2\left( {x + 4} \right)\left( {x + 5} \right) + 3\left( {x + 3} \right)\left( {x + 5} \right) + 4\left( {x + 3} \right)\left( {x + 4} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right)}}} \right]$ $\frac{{dy}}{{dx}} = y\left[ {\frac{{2\left( {{x^2} + 9x + 20} \right) + 3\left( {{x^2} + 8x + 15} \right) + 4\left( {{x^2} + 7x + 12} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right)}}} \right]$ $= \frac{{{{\left( {x + 3} \right)}^2} \cdot {{\left( {x + 4} \right)}^3} \cdot {{\left( {x + 5} \right)}^4}}}{{\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right)}}\left[ {2{x^2} + 18x + 40 + 3{x^2} + 24x + 45 + 4{x^2} + 28x + 48} \right]$ $= \left( {x + 3} \right){\left( {x + 4} \right)^2}{\left( {x + 5} \right)^3}\left( {9{x^2} + 70x + 133} \right)$

Question (6)

${\left( {x + \frac{1}{x}} \right)^x} + {x^{\left( {1 + \frac{1}{x}} \right)}}$

Solution

Here variable raised to variable is given so we have to take log on both sides, but log(a+b) does not have any property. So we assume each one as new variable and then solve separately $\log {a^m} = m\log a$ $\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$ $\frac{d}{{dx}}{a^x} = {a^x}\log a$ $\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ $\frac{d}{{dx}}\left( {\frac{1}{x}} \right) = \frac{{ - 1}}{{{x^2}}}$
$\text{Let} \quad u = {\left( {x + \frac{1}{x}} \right)^x}$ Take log on both sides $\log u = x\log \left( {x + \frac{1}{x}} \right)$ Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log \left( {x + \frac{1}{x}} \right) + \log \left( {x + \frac{1}{x}} \right)\frac{d}{{dx}}x$ $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{1}{{\left( {x + \frac{1}{x}} \right)}}\frac{d}{{dx}}\left( {x + \frac{1}{x}} \right) + \log \left( {x + \frac{1}{x}} \right) \cdot \left( 1 \right)$ $\frac{{du}}{{dx}} = u\left[ {x\frac{x}{{\left( {{x^2} + 1} \right)}}\left( {1 - \frac{1}{{{x^2}}}} \right) + \log \left( {x + \frac{1}{x}} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {x + \frac{1}{x}} \right)^x}\left[ {\frac{{{x^2}}}{{{x^2} + 1}} \cdot \frac{{{x^2} - 1}}{{{x^2}}} + \log \left( {x + \frac{1}{x}} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {x + \frac{1}{x}} \right)^x}\left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}} + \log \left( {x + \frac{1}{x}} \right)} \right]$ $\text{Let} \quad v = {\left( x \right)^{1 + \frac{1}{x}}}$ Take log on both sides $\log v = \log \left[ {{{\left( x \right)}^{1 + \frac{1}{x}}}} \right]$ $\log v = \left( {1 + \frac{1}{x}} \right)\log x$ Differentiate w.r.t x $\frac{1}{v}\frac{{dv}}{{dx}} = \left( {1 + \frac{1}{x}} \right)\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}\left( {1 + \frac{1}{x}} \right)$ $\frac{1}{v}\frac{{dv}}{{dx}} = \left( {1 + \frac{1}{x}} \right)\frac{1}{x} + \log x\left( { - \frac{1}{{{x^2}}}} \right)$ $\frac{1}{v}\frac{{dv}}{{dx}} = \frac{{x + 1}}{x} \cdot \frac{1}{x} - \frac{{\log x}}{{{x^2}}}$ $\frac{{dv}}{{dx}} = v\left[ {\frac{{x + 1}}{{{x^2}}} - \frac{{\log x}}{{{x^2}}}} \right]$ $\frac{{dv}}{{dx}} = {\left( x \right)^{1 + \frac{1}{x}}}\left[ {\frac{{x + 1 - \log x}}{{{x^2}}}} \right]$ $\text{now} \quad y = {\left( {x + \frac{1}{x}} \right)^x} + {\left( x \right)^{1 + \frac{1}{x}}}$ y = u + v Differentiate w.r.t x
$\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ $\frac{{dy}}{{dx}} = \left( {x + \frac{1}{x}} \right)\left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}} + \log \left( {x + \frac{1}{x}} \right)} \right] + {x^{\frac{1}{x} - 1}}\left( {x + 1 - \log x} \right)$

Question (7)

(log x)x + xlogx

Solution

Here variable raised to variable is given so we have to take log on both sides, but log(a+b) does not have any property. So we assume each one as new variable and then solve separately $\log {a^m} = m\log a$ $\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$ $\frac{d}{{dx}}{a^x} = {a^x}\log a$ $\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ $\frac{d}{{dx}}\left( {\frac{1}{x}} \right) = \frac{{ - 1}}{{{x^2}}}$
Let u = (logx)x
Taking log on both side
log u = log[(logx)x]
logu = xlog(logx)
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log \left( {\log x} \right) + \log \left( {\log x} \right)\frac{d}{{dx}}x$ $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{1}{{\log x}}\frac{1}{x} + \log \left( {\log x} \right) \cdot 1$ $\frac{{du}}{{dx}} = u\left[ {\frac{1}{{\log x}} + \log \left( {\log x} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {\log x} \right)^x}\left[ {\frac{{1 + \log x \cdot \log \left( {\log x} \right)}}{{\log x}}} \right]$ $\frac{{du}}{{dx}} = {\left( {\log x} \right)^{x - 1}}\left[ {1 + \log x \cdot \log \left( {\log x} \right)} \right]$ Let v = xlogx
Take log on both sides
logv = log[xlogx]
logv=logx.(logx) = (logx)2
Differentiate w.r.t x $\frac{1}{v}\frac{{dv}}{{dx}} = 2\left( {\log x} \right)\frac{d}{{dx}}\log x$ $\frac{1}{v}\frac{{dv}}{{dx}} = \frac{{2\log x}}{x}$ $\frac{{dv}}{{dx}} = v\left[ {\frac{{2\log x}}{x}} \right]$ $\frac{{dv}}{{dx}} = {x^{\log x}}\left[ {\frac{{2\log x}}{x}} \right]$ $\frac{{dv}}{{dx}} = {x^{\log x - 1}}\left( {2\log x} \right)$ $\frac{{dv}}{{dx}} = 2{x^{\log x - 1}}\log x$ Now y = (log x)x + xlogx
y = u +v
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ $\frac{{dy}}{{dx}} = {\left( {\log x} \right)^{x - 1}}\left[ {1 + \log x\log \left( {\log x} \right)} \right] + 2{x^{\log x - 1}}\log x$

Question (8)

(sin x)x + sin-1 √x

Solution

Here variable raised to variable is given so we have to take log on both sides, but log(a+b) does not have any property. So we assume each one as new variable and then solve separately $\log {a^m} = m\log a$ $\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$ $\frac{d}{{dx}}{a^x} = {a^x}\log a$ $\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$ $\frac{d}{{dx}}\left( {\frac{1}{x}} \right) = \frac{{ - 1}}{{{x^2}}}$ $\frac{d}{{dx}}{\sin ^{ - 1}}x = \frac{1}{{\sqrt {1 - {x^2}} }}$
Let u = (sinx)x
Take log on both side log u = log[ (sinx)x]
log u=xlog(sinx)
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log \left( {\sin x} \right) + \log \left( {\sin x} \right)\frac{d}{{dx}}x$ $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{1}{{\sin x}}\cos x + \log \left( {\sin x} \right)$ $\frac{1}{u}\frac{{du}}{{dx}} = x\cot x + \log \left( {\sin x} \right)$ $\frac{{du}}{{dx}} = u\left[ {x\cot x + \log \left( {\sin x} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {x\cot x + \log \left( {\sin x} \right)} \right]$ y= (sinx)x + sin-1√x
y= u + sin-1√x
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{1}{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}\frac{d}{{dx}}\sqrt x$ $\frac{{dy}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {x\cot x + \log \left( {\sin x} \right)} \right] + \frac{1}{{\sqrt {1 - x} }}\frac{1}{{2\sqrt x }}$ $\frac{{dy}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {x\cot x + \log \left( {\sin x} \right)} \right] + \frac{1}{{2\sqrt {x - {x^2}} }}$

Question (9)

xsinx + (sin x)cosx

Solution

Let u= xsinx
Taking log on both sides
log u = log(xsinx)
log u = sinx logx
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = \sin x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}\sin x$ $\frac{{du}}{{dx}} = u\left[ {\frac{{\sin x}}{x} + \log x \cdot \cos x} \right]$ $\frac{{du}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x}}{x} + \log x \cdot \cos x} \right]$ Let v = (sinx)cosx
Taking log on both sides
log v = log[(sinx)cosx]
log v = cosx .log(sinx)
Differentiate w.r.t x $\frac{1}{v}\frac{{dv}}{{dx}} = \cos x\frac{d}{{dx}}\log \left( {\sin x} \right) + \log \left( {\sin x} \right)\frac{d}{{dx}}\cos x$ $\frac{1}{v}\frac{{dv}}{{dx}} = \cos x\frac{1}{{\sin x}}\frac{d}{{dx}}\sin x + \log \left( {\sin x} \right)\left( { - \sin x} \right)$ $\frac{1}{v}\frac{{dv}}{{dx}} = \frac{{\cos x}}{{\sin x}}\cos x - \sin x\log \left( {\sin x} \right)$ $\frac{{dv}}{{dx}} = v\left[ {\cot x\cos x - \sin x\log \left( {\sin x} \right)} \right]$ $\frac{{dv}}{{dx}} = {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \sin x\log \left( {\sin x} \right)} \right]$ Now y = xsinx + (sinx)cosx
y = u + v
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ $\frac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\frac{{\sin x}}{x} + \log x\cos x} \right] + {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \sin x\log \left( {\sin x} \right)} \right]$

Question (10)

${x^{x\cos x}} + \frac{{{x^2} + 1}}{{{x^2} - 1}}$

Solution

Let u = xxcosx
Take log on both side
log u = xcosx logx
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\cos x\frac{d}{{dx}}\log x + x\log x\frac{d}{{dx}}\cos x + \cos x\log x\frac{d}{{dx}}x$ $\frac{{du}}{{dx}} = u\left[ {\require{cancel} \cancel{x}\cos x \cdot \frac{1}{\cancel{x}} + x\log x\left( { - \sin x} \right) + \cos x\log x} \right]$ $\frac{{du}}{{dx}} = {x^{x\cos x}}\left[ {\cos x + \cos x\log x - x\log x\sin x} \right]$ $\frac{{du}}{{dx}} = {x^{x\cos x}}\left[ {\cos x\left( {1 + \log x} \right) - x\log x\sin x} \right]$ $v = \frac{{{x^2} + 1}}{{{x^2} - 1}}$ Differentiate w.r.t x $\frac{{dv}}{{dx}} = \frac{{\left( {{x^2} - 1} \right)\frac{d}{{dx}}\left( {{x^2} + 1} \right) - \left( {{x^2} + 1} \right)\frac{d}{{dx}}\left( {{x^2} - 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ $\frac{{dv}}{{dx}} = \frac{{\left( {{x^2} - 1} \right)\left( {2x} \right) - \left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ $\frac{{dv}}{{dx}} = \frac{{2x\left[ {\cancel{{x^2}} - 1 - \cancel{{x^2}} - 1} \right]}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ $\frac{{dv}}{{dx}} = \frac{{ - 4x}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ $\text{Now} \quad y = {x^{x\cos x}} + \frac{{{x^2} - 1}}{{{x^2} + 1}}$ y = u + v
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ $\frac{{dy}}{{dx}} = {x^{x\cos x}}\left[ {\cos x\left( {1 + x\log x} \right) - x\sin x\log x} \right] - \frac{{4x}}{{{{\left( {{x^2} - 1} \right)}^2}}}$

Question (11)

(x cosx)x) + (xsinx)1/x

Solution

y = (xcosx)x) + (xsinx)1/x
Let u = (xcosx)x
Take log on both sides
log u = log [(xcosx)x]
log u = xlog(xcosx)
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = x\frac{d}{{dx}}\log \left( {x\cos x} \right) + \log \left( {x\cos x} \right)\frac{d}{{dx}}x$ $\frac{1}{u}\frac{{du}}{{dx}} = x\left[ {\frac{1}{{x\cos x}}\left( {x\left( { - \sin x} \right) + \cos x\left( 1 \right)} \right)} \right] + \log \left( {x\cos x} \right)$ $\frac{1}{u}\frac{{du}}{{dx}} = \frac{{ - x\sin x + \cos x}}{{\cos x}} + \log \left( {x\cos x} \right)$ $\frac{{du}}{{dx}} = u\left[ {\frac{{ - x\sin x + \cos x}}{{\cos x}} + \log \left( {x\cos x} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ { - x\tan x + 1 + \log \left( {x\cos x} \right)} \right]$ $\frac{{du}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$ Let V = (xsinx)1/x
Take log on both side
log v = log(xsinx)1/x $\log v = \frac{1}{x}\log \left( {x\sin x} \right)$ Differentiate w.r.t x $\frac{1}{v}\frac{{dv}}{{dx}} = \frac{1}{x}\frac{d}{{dx}}\log \left( {x\sin x} \right) + \log \left( {x\sin x} \right) \cdot \frac{d}{{dx}}\left( {\frac{1}{x}} \right)$ $\frac{1}{v}\frac{{dv}}{{dx}} = \frac{1}{x}\frac{1}{{x\sin x}}\left[ {x\cos x + \sin x} \right] + \log \left( {x\sin x} \right) \times \frac{{ - 1}}{{{x^2}}}$ $\frac{{dv}}{{dx}} = v\left[ {\frac{{x\cos x + \sin x}}{{{x^2}\sin x}} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$ $\frac{{dv}}{{dx}} = {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{{x\cot x + 1 - \log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$ Now y = u + v
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ $\frac{{dy}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] + {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{{x\cot x + 1 - \log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$

### Find dy/dx of the function given in Exercises 12 to 15

Question (12)

xy + yx = 1

Solution

Let u = xy
log u = ylogx
Differentiate w.r.t x $\frac{1}{u}\frac{{du}}{{dx}} = y\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}y$ $\frac{{du}}{{dx}} = u\left[ {\frac{y}{x} + \log x\frac{{dy}}{{dx}}} \right]$ $\frac{{du}}{{dx}} = {x^y}\left[ {\frac{y}{x} + \log x\frac{{dy}}{{dx}}} \right]$ $\frac{{du}}{{dx}} = y{x^{y - 1}} + {x^y}\log x\frac{{dy}}{{dx}}$ Let yx = v
Take log on both side
log v = x log y
Differentiate w.r.t x $\frac{1}{v}\frac{{dv}}{{dx}} = x\frac{d}{{dx}}\log y + \log y\frac{d}{{dx}}x$ $\frac{{dv}}{{dx}} = v\left[ {x\frac{1}{y}\frac{{dy}}{{dx}} + \log y} \right]$ $\frac{{dv}}{{dx}} = {y^x}\left[ {\frac{x}{y}\frac{{dy}}{{dx}} + \log y} \right]$ $\frac{{dv}}{{dx}} = x{y^{x - 1}}\frac{{dy}}{{dx}} + {y^x}\log y$ Now xy + yx = 1
u + v = 1
Differentiate w.r.t x $\frac{{du}}{{dx}} + \frac{{dv}}{{dx}} = 0$ $y{x^{y - 1}} + {x^y}\log x\frac{{dy}}{{dx}} + x{y^{x - 1}}\frac{{dy}}{{dx}} + {y^x}\log y = 0$ $\frac{{dy}}{{dx}}\left[ {{x^y}\log x + x{y^{x - 1}}} \right] = - \left[ {{y^x}\log y + y{x^{y - 1}}} \right]$ $\frac{{dy}}{{dx}} = \frac{{ - \left[ {{y^x}\log y + y{x^{y - 1}}} \right]}}{{\left[ {{x^y}\log x + x{y^{x - 1}}} \right]}}$

Question (13)

yx = xy

Solution

Take log on both side xlog y = y logx
Differentiate w.r.t x $x\frac{d}{{dx}}\log y + \log y\frac{d}{{dx}}x = y\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}y$ $x\frac{1}{y}\frac{{dy}}{{dx}} + \log y = \frac{y}{x} + \log x\frac{{dy}}{{dx}}$ $\therefore \frac{x}{y}\frac{{dy}}{{dx}} - \log x\frac{{dy}}{{dx}} = \frac{y}{x} - \log y$ $\frac{{dy}}{{dx}}\left[ {\frac{x}{y} - \log x} \right] = \left[ {\frac{y}{x} - \log x} \right]$ $\frac{{dy}}{{dx}} = \frac{y}{x}\frac{{\left( {y - x\log y} \right)}}{{\left( {x - y\log x} \right)}}$

Question (14)

(cos x)y = (cos y)x

Solution

Take log on both side
ylog(cosx) = xlog(cosy)
Differentiate w.r.t x $y\frac{d}{{dx}}\log \left( {\cos x} \right) + \log \left( {\cos x} \right)\frac{d}{{dx}}y = x\frac{d}{{dx}}\log \left( {\cos y} \right) + \log \left( {\cos y} \right)\frac{d}{{dx}}x$ $y\frac{1}{{\cos x}} \times - \sin x + \log \left( {\cos x} \right)\frac{{dy}}{{dx}} = \frac{x}{{\cos y}} \times - \sin y\frac{{dy}}{{dx}} + \log \left( {\cos y} \right) \cdot 1$ $- y\tan x + \log \left( {\cos x} \right)\frac{{dy}}{{dx}} = - x\tan y\frac{{dy}}{{dx}} + \log \left( {\cos y} \right)$ $\log \left( {\cos x} \right)\frac{{dy}}{{dx}} + x\tan y\frac{{dy}}{{dx}} = \log \left( {\cos y} \right) + y\tan x$ $\log \left( {\cos x} \right)\frac{{dy}}{{dx}} + x\tan y\frac{{dy}}{{dx}} = \log \left( {\cos y} \right) + y\tan x$ $\frac{{dy}}{{dx}}\left[ {\log \left( {\cos x} \right) + x\tan y} \right] = \log \left( {\cos y} \right) + y\tan x$ $\frac{{dy}}{{dx}} = \frac{{\log \left( {\cos y} \right) + y\tan x}}{{\log \left( {\cos x} \right) + x\tan y}}$

Question (15)

xy = e(x-y)

Solution

Take log on both side
log(xy) = log ex-y
logx+logy = (x-y).loge
logx+logy = (x-y)
y + logy = x - logx
Differentiate w.r.t x $\frac{{dy}}{{dx}} + \frac{1}{y}\frac{{dy}}{{dx}} = 1 - \frac{1}{x}$ $\frac{{dy}}{{dx}}\left( {1 + \frac{1}{y}} \right) = 1 - \frac{1}{x}$ $\frac{{dy}}{{dx}}\left( {\frac{{y + 1}}{y}} \right) = \frac{{x - 1}}{x}$ $\frac{{dy}}{{dx}} = \frac{{y\left( {x - 1} \right)}}{{x\left( {y + 1} \right)}}$

Question (16)

Find the derivative of function given by
f(x) = (1+x)(1+x2)(1+x4) (1+x5)
and hence find f'(1)

Solution

Take log on both sides
log f(x) = log(1+x) + log(1+x2) + log(1+x4) + log(1+x8)
Differentiate w.r.t x $\frac{1}{{f\left( x \right)}} \cdot f'\left( x \right) = \frac{1}{{1 + x}} + \frac{1}{{1 + {x^2}}}2x + \frac{1}{{1 + {x^4}}}4{x^3} + \frac{1}{{1 + {x^8}}}8{x^7}$ $f'\left( x \right) = f\left( x \right)\left[ {\frac{1}{{1 + x}} + \frac{1}{{1 + {x^2}}}2x + \frac{1}{{1 + {x^4}}}4{x^3} + \frac{1}{{1 + {x^8}}}8{x^7}} \right]$ $f'\left( x \right) = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right)\left[ {\frac{1}{{1 + x}} + \frac{{2x}}{{1 + {x^2}}} + \frac{{4{x^3}}}{{1 + {x^4}}} + \frac{{8{x^7}}}{{1 + {x^8}}}} \right]$ value of f(1) $f\left( 1 \right) = \left( 2 \right)\left( 2 \right)\left( 2 \right)\left( 2 \right) = 16$ value of f'(1) $f'\left( x \right) = 16\left[ {\frac{1}{{1 + 1}} + \frac{{2\left( 1 \right)}}{{1 + 1}} + \frac{{4\left( 1 \right)}}{{1 + 1}} + \frac{{8\left( 1 \right)}}{{1 + 1}}} \right]$ $f'\left( 1 \right) = 16\left[ {\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}} \right]$ $f'\left( 1 \right) = \cancel{16}^8 \times \frac{{15}}{\cancel{2}}$ f'(1) = 120

Question (17)

Differentiate (x2 -5x+8) (x3+7x+9) in three ways mentioned below
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial
(iii) by logarithmic differentiation
Do they all give the same answer?

Solution

(i) By product rule
y=(x2 - 5x +8) (x3 + 7x + 9)
Differentiate w.r.t x $\frac{{dy}}{{dx}} = \left( {{x^2} - 5x + 8} \right)\frac{d}{{dx}}\left( {{x^3} + 7x + 9} \right) + \left( {{x^3} + 7x + 9} \right)\frac{d}{{dx}}\left( {{x^2} - 5x + 8} \right)$ $\frac{{dy}}{{dx}} = \left( {{x^2} - 5x + 8} \right)\left( {3{x^2} + 7} \right) + \left( {{x^3} + 7x + 9} \right)\left( {2x - 5} \right)$ $= 3{x^4} + 7{x^2} - 15{x^3} - 35x + 24{x^2} + 56 + 2{x^4} - 5{x^3} + 14{x^2} - 35x + 18x - 45$ $= 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11$ (ii) By expanding the product
y= (x2 - 5x + 8) (x3 + 7x + 9)
y = x5 + 7x3 + 9x2 - 5x4 - 35x2 -45x +8x3 +56x+72
y = x5 -5x4 +15x3 -26x2 + 11x +72
Differentiate w.r.t x $\frac{{dy}}{{dx}} = 5{x^4} - 5\left( {4{x^3}} \right) + 15\left( {3{x^2}} \right) - 26\left( {2x} \right) + 11\left( 1 \right) + 0$ $\frac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11$ (iii) By log
y =(x2 -5x+8) (x3+7x+9)
Take log on both side
log y = log(x2 -5x+8) + log (x3+7x+9)
Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{{x^2} - 5x + 8}} \times \left( {2x - 5} \right) + \frac{1}{{{x^3} + 7x + 9}} \times \left( {3{x^2} + 7} \right)$ $\frac{{dy}}{{dx}} = y\left[ {\frac{{2x - 5}}{{{x^2} - 5x + 8}} + \frac{{3{x^2} + 7}}{{{x^3} + 7x + 9}}} \right]$ $= \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)\left[ {\frac{{\left( {2x - 5} \right)\left( {{x^3} + 7x + 9} \right) + \left( {3{x^2} + 7} \right)\left( {{x^2} - 5x + 8} \right)}}{{\left( {{x^3} + 7x + 9} \right)\left( {{x^2} - 5x + 8} \right)}}} \right]$ =2x4+14x2+18x-5x3-35x-45+3x4-15x3+24x2 +7x2-35x+56
=5x4 -20x3+45x2-52x+11
Yes, all will give same answer

Question (18)

If u, v and w are function of x, then show that $\frac{d}{{dx}}\left( {u,v,w} \right) = \frac{{du}}{{dx}}v \cdot w + u \cdot \frac{{dv}}{{dx}} \cdot w + u \cdot v\frac{{dw}}{{dx}}$ in two ways-first by repeated application of product rule, second vy logarithmic differentiation

Solution

(i) Repeated application of product rule
y = [uvw] = u[vw]
Differentiate w.r.t x $\frac{{dy}}{{dx}} = u\frac{d}{{dx}}\left( {vw} \right) + vw\frac{d}{{dx}}u$ $\frac{{dy}}{{dx}} = u\left[ {v\frac{{dw}}{{dx}} + w\frac{{dv}}{{dx}}} \right] + vw\frac{{du}}{{dx}}$ $\frac{{dy}}{{dx}} = uv\frac{{dw}}{{dx}} + uw\frac{{dv}}{{dx}} + vw\frac{{du}}{{dx}}$ $\frac{{dy}}{{dx}} = \frac{{du}}{{dx}}vw + u\frac{{dv}}{{dx}}w + uv\frac{{dw}}{{dx}}$ (ii) By log
y = u v w
Take log on both side
log y = log (u.v.w)
log y = logu + logv + logw
Differentiate w.r.t x $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}$ $\frac{{dy}}{{dx}} = y\left[ {\frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}} \right]$ $\frac{{dy}}{{dx}} = uvw\left[ {\frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}} \right]$ $\frac{{dy}}{{dx}} = vw\frac{{du}}{{dx}} + uw\frac{{dv}}{{dx}} + uv\frac{{dw}}{{dx}}$