Chapter 4.1, Problem 39E

To determine

To calculate: The $dy/dx\text{for}y=\cos (6x+2)$ .

Answer to Problem 39E

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) $y=\cos u,u=6x+2$ and b) $y=\cos 2u,u=3x+1$

Formula used:

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

Also quotient, chain and addition rule.

Calculation:

If $y=\cos u,u=6x+2$

Then,

  $y=f\left(g\left(x\right)\right)=\cos \left(6x+2\right)$

Differentiating the function with x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d(\cos (6x+2))}{dx}\\ =\left(-\sin \left(6x+2\right)\right)\times 6\left(1\right)\\ =-6\sin \left(6x+2\right)\end{array}$

Now if $y=\cos 2u,u=3x+1$

Then,

  $y=f\left(g\left(x\right)\right)=\cos \left(6x+2\right)$

Differentiating the function with x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d(\cos 2(3x+1))}{dx}\\ =\left(-\sin 2\left(3x+1\right)\times 2\right)\times 3\left(1\right)\\ =-6\sin \left(6x+2\right)\end{array}$

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.