Question

Differentiate the following :
$\sf (i) \: \: {x}^{ x } + 2^{\sin x} \\ \\ \sf (ii) {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } \: \: using \: parametric \: form$​

Answer 1

Solution:–

1)

$let \: y = {x}^{x} - {2}^{sinx}$

$also \: let \: {x}^{x} = u \: and \: {2}^{sinx} = v$

$\therefore \: y = u - v$

$\implies \: \frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$

$u = {x}^{x}$

Talking logarithm on both sides

$log(u) = x log(x)$

Differentiating both sides with respect to x

$\frac{1}{u} . \frac{du}{dx} = ( \frac{d}{dx} (x) \times log(x) + x \times \frac{d}{dx} \times log(x) )$

$\implies \frac{du}{dx} = u(1 \times log(x) + x \times \frac{1}{x} )$

$\implies \: \frac{du}{dx} = {x}^{ x} (1 + log(x) )$

$v = {2}^{sinx}$

Taking logarithm on both sides with respect to x

differentiating,

$\frac{1}{v} . \frac{dv}{dx} = log(2) . \frac{d}{dx} ( \sin(x) )$

$\implies \: \frac{dv}{dx} = v log(2) \cos(x)$

$\implies \: \frac{dv}{dx} = {2}^{sinx} \cos(x) log(2)$

$\therefore \: \frac{dy}{dx} = {x}^{x} (1 + log(x) - {2}^{sinx} cosx$log2

2)

${x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} }$

differentiating w.r.t.x

$\frac{2}{3} . {x}^{ \frac{2}{3} - 1 } + \frac{2}{3} . {y}^{ \frac{2}{3} - 1 } \: \: \: \: \: \: \: \: \: \: \frac{dy}{dx} = 0(because \: a \: is \: constant)$

now,

$\frac{2}{3} {x}^{ \frac{ - 1}{3} } + \frac{2}{ {3y}^{ - \frac{1}{3} } } \: \: \: \: \: \: \: \frac{dy}{dx} = 0$

$\frac{dy}{dx} = - {( \frac{x}{y}) }^{ \frac{ - 1}{3} }$

Answer 2

ANSWER 1:

$x^x (1+log x) +2^{sin x}cos x \: log 2$

FORMULAE USED:

$\log {x}^{y} = y \log x$

$\dfrac{d}{dx} (log \: x) = \dfrac{1}{x}$

$\dfrac{d}{dx}(x) = 1$

$\dfrac{d}{dx} ( \sin x) = \cos x$

$\dfrac{d}{dx} (constant) = 0$

EXPLANATION:

REFER ATTACHMENT FOR EXPLANATION.

ANSWER 2:

$\dfrac{ dy}{dx} = - \sqrt[3]{ \dfrac{y}{x} }$

EXPLANATION:

Here powers are given as 2/3 so we consider

x = a sin³ θ and y = a cos³ θ

This will help as eliminate the cube root ie power(1/3)

${(a \sin^{3} \theta)}^{ \frac{2}{3} } + {(a \cos^{3} \theta)}^{ \frac{2}{3} }$

$a^{2/3} \sin^2\theta+a^{2/3} \cos^2\theta$

$a^{2/3} (\sin^2\theta+ \cos^2\theta) = a^{2/3}$

$\dfrac{dy}{dx} = \frac{ \dfrac{dy}{d \theta} }{ \dfrac{dx}{d \theta} }$

$\dfrac{dy}{ d \theta} = a(3 { \cos}^{2} \theta) - \sin \theta$

$\dfrac{dy}{d \theta} = -3 a \sin \theta { \cos}^{2} \theta$

$\dfrac{dx}{d \theta} = a(3 { \sin}^{2} \theta) \cos \theta$

$\dfrac{dx}{d \theta} = 3a{ \sin}^{2} \theta \cos \theta$

$\dfrac{ dy}{dx} = \dfrac{-3 a \sin \theta { \cos}^{2} \theta}{3a{ \sin}^{2} \theta \cos \theta}$

$\dfrac{ dy}{dx} = - \dfrac{ \cos \theta}{ \sin \theta}$

$\cos \theta = {y}^{ \frac{1}{3} }$

$\sin \theta = {x}^{ \frac{1}{3} }$

$\dfrac{ dy}{dx} = - \bigg( { \dfrac{y}{x}\bigg) }^{ \frac{1}{3} }$

$\dfrac{ dy}{dx} = - \sqrt[3]{ \dfrac{y}{x} }$