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  • Differentiate each of the following w.r.t. x cos(tan root x plus 1)

Differentiate each of the following w.r.t. x
\cos (\tan \sqrt{x+1})

Answers (1)

We have given  \cos (\tan \sqrt{x+1})

Let us Assume \sqrt{x+1}=w
And \tan \sqrt{x+1}=v
\mathrm{So}, \mathrm{y}=\cos \mathrm{v}
Now, differentiate w.r.t v
\frac{\mathrm{dy}}{\mathrm{d} \mathrm{v}}=(-\sin \mathrm{v})
And, \mathrm{v}= tan \mathrm{w}
Now, again differentiate w.r.t. w
\frac{d v}{d w}=\sec ^{2} w

And, we know, \sqrt{x+1}=w
So, differentiate w w.r.t. x we get
\frac{d w}{d x}=\frac{1}{2 \sqrt{x+1}}
Now, using chain rule we get,
\\\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{dw}} \times \frac{\mathrm{d} \mathrm{w}}{\mathrm{dx}}$ \\$\frac{d y}{d x}=(-\sin v) \times \sec ^{2} w \times \frac{1}{2 \sqrt{x+1}}
Substitute the value of v and w
\frac{\mathrm{dy}}{\mathrm{dx}}=\left(-\sin (\tan \sqrt{\mathrm{x}+1}) \times \sec ^{2} \sqrt{\mathrm{x}+1} \times \frac{1}{2 \sqrt{\mathrm{x}+1}}\right).

Hence, dy/dy is the differentiation of function.

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infoexpert22

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