Chapter 4.1, Problem 40E

To determine

To calculate:And compare the $dy/dx\text{for}y=\sin \left(x^2+1\right)$ $\text{by}\text{writing}\text{y}\text{as}\text{a}\text{composite}\text{with}$

  $\begin{array}{l}\text{a}\text{.}y=\sin \left(u+1\right),u=x^2\\ \text{b}\text{.}y=\sin u,u=x^2+1\end{array}$ .

Answer to Problem 40E

Hence it can be seen that if a function y is written as a composite in different ways then in both the variation the $dy/dx$ of function y is coming the same.

Explanation of Solution

Given information: Given function are a) $y=\sin \left(u+1\right),u=x^2$ and b) $y=\sin u,u=x^2+1$

Formula used:

  $\begin{array}{l}\frac{d(f(g(x)))}{dx}=f\text{'}(g(x))\times g\text{'}(x)\\ \frac{d(\sin x)}{dx}=\cos x\\ \frac{d(x^n)}{dx}=n{x}^{n-1}\end{array}$

Also quotient, chain and addition rule.

Calculation:

If $y=\sin \left(u+1\right),u=x^2$

Then,

  $y=f\left(g\left(x\right)\right)=\sin \left(x^2+1\right)$

Differentiating the function w.r.t x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d\left(\sin \left(\left(x^2\right)+1\right)\right)}{dx}\\ =\left(\cos \left(x^2+1\right)\times \left(1\right)\right)\times 2\left(x\right)\\ =2x\cos \left(x^2+1\right)\end{array}$

Now if $y=\sin u,u=x^2+1$

Then,

  $y=f\left(g\left(x\right)\right)=\sin \left(x^2+1\right)$

Differentiating the function w.r.t x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d\left(\sin \left(x^2+1\right)\right)}{dx}\\ =\cos \left(x^2+1\right)\times \left(2x\right)\\ =2x\cos \left(x^2+1\right)\end{array}$

Hence it can be seen that if a function y is written as a composite in different ways then in both the variation the $dy/dx$ of function y is coming the same.