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Chapter 6: Problem 81

Find \(\frac{d y}{d x}\), if \(y=x^{\sin x-\cos x}+\frac{x^{2}-1}{x^{2}+1}\).

Short Answer

Expert verified
The derivative of \(y=x^{\sin x-\cos x}+\frac{x^{2}-1}{x^{2}+1}\) is \(x^{\sin x-\cos x} * ((\cos x + \sin x) * \ln(x) + (\sin x - \cos x)/x) + \frac{4x}{(x^{2}+1)^{2}}\).

Step by step solution

01

Differentiate First Term

First differentiate \(x^{\sin x-\cos x}\). This will involve the use of the chain rule and power rule. The derivative of \(u^v\) where both u and v are functions of x is given by \(u^v*(v' * \ln(u) + \frac{v*u'}{u})\). Here, \(u = x\) and \(v = \sin x - \cos x\). So, find the derivatives \(u'\) and \(v'\), which are \(1\) and \(\cos x + \sin x\) respectively.
02

Apply the Derivation Rule to the First Term

Now, putting these values into our derivative rule, we get the derivative of \(x^{\sin x-\cos x}\) as \(x^{\sin x-\cos x} * ((\cos x + \sin x) * \ln(x) + \frac{(\sin x - \cos x) * 1}{x})\).
03

Differentiate Second Term

Now find the derivative of \(\frac{x^{2}-1}{x^{2}+1}\). This will require the quotient rule, which is \((v*u' - u*v')/v^{2}\) for a function of the form \(\frac{u}{v}\). Here \(u=x^{2}-1\) and \(v=x^{2}+1\), and their derivatives \(u'\) and \(v'\) are \(2x\) and \(2x\) respectively.
04

Apply the Derivation Rule to the Second Term

Applying the quotient rule, we get the derivative \(\frac{(x^{2}+1)*2x - (x^{2}-1)*2x}{(x^{2}+1)^{2}} = \frac{4x}{(x^{2}+1)^{2}}\).
05

Combine the Derivatives

To finish, combine the derivatives from step 2 and step 4. This gives the total derivative of the original function: \(x^{\sin x-\cos x} * ((\cos x + \sin x) * \ln(x) + (\sin x - \cos x)/x) + \frac{4x}{(x^{2}+1)^{2}}\).

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Most popular questions from this chapter

If \(y=x \sin y\), prove that \(\frac{d y}{d x}=\frac{y}{x(1-x \cos y)} .\)

If \(y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right)\) \(0

If \(f(x)=a x+b, x \in[-2,2]\), then the point \(c \in(-2,2)\) where $$ f(c)=\frac{f(2)-f(-2)}{4} $$ (a) does not exist (b) can be any \(c \in(-2,2)\) (c) can be only 1 (d) can be only \(-1\).

If \(y=\tan ^{-1} x\), then prove that (i) \(\left(1+x^{2}\right) y_{2}+2 x y_{1}=0\) (ii) \(\left(1+x^{2}\right) y_{n+2}+2(n+2) x y_{n+1}+n(n+1) y_{n}\) \(=0 .\)

If \(y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1\) prove that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
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