Answer to Question #91411 in Calculus for Fariha

Question

Find the rate of change of h(x)=2 cos⁡〖(3x+tan⁡x)〗 with respect to x.

Answer 1

The rate of change of y with respect to x, if one has the original function, can be found by taking the derivative of that function. This will measure the rate of change at a specific point.

$$h(x)=2 \cos⁡(3x+\tan⁡(x))$$

The derivative of f with respect to x,$h(x)'$ is given by

$$h'(x)=(2 \cos⁡(3x+\tan⁡(x)))^{\prime}= \\ =-2\sin(3x+\tan⁡(x))(3x+\tan⁡(x))^{\prime}=\\ =-2\sin(3x+\tan⁡(x))(3+\frac{1}{\cos^2(x)})$$

So,

$$h'(x)=-2\sin(3x+\tan⁡(x))\frac{3\cos^2(x)+1}{\cos^2(x)}=-\frac{2\sin(3x+\tan⁡(x))(3\cos^2(x)+1)}{\cos^2(x)}$$

We found the rate of change of $h(x)=2 \cos⁡(3x+\tan⁡(x))$ with respect to x: 

$$h'(x)=-\frac{2\sin(3x+\tan⁡(x))(3\cos^2(x)+1)}{\cos^2(x)}$$