Multivariable Chain Rule – Exercise 3329

Exercise

Given

$$z(t,x,y)=\tan (3t+2x^2-y)$$

$$f(t)=\frac{1}{t}$$

$$v(t)=\sqrt{t}$$

$$Z(t)=z(t,f(t),v(t))$$

Calculate the derivative

$$Z'(t)$$

$$Z'(t)=\frac{3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}}}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}$$

Solution

We will use the chain rule to calculate the derivative

$$Z'(t)=z'_t+z'_f\cdot f'_t+z'_v\cdot v'_t$$

We have

$$Z(t)=z(t,f,v)$$

And function z is

$$z(t,x,y)=\tan (3t+2x^2-y)$$

Hence, we get

$$z(t,f,v)=\tan (3t+2f^2-v)$$

We calculate the partial derivatives of z.

$$z'_t=\frac{1}{\cos^2(3t+2f^2-v)}\cdot 3=$$

$$=\frac{3}{\cos^2(3t+2f^2-v)}$$

$$z'_f=\frac{1}{\cos^2(3t+2f^2-v)}\cdot 4f=$$

$$=\frac{4f}{\cos^2(3t+2f^2-v)}$$

$$z'_v=\frac{1}{\cos^2(3t+2f^2-v)}\cdot (-1)=$$

$$=\frac{-1}{\cos^2(3t+2f^2-v)}$$

And the partial derivatives of f and v.

$$f'_t=\frac{-1}{t^2}$$

$$v'_t=\frac{1}{2\sqrt{t}}$$

We put the derivatives and get

$$Z'(t)=z'_t+z'_f\cdot f'_t+z'_v\cdot v'_t=$$

$$=\frac{3}{\cos^2(3t+2f^2-v)}+\frac{4f}{\cos^2(3t+2f^2-v)}\cdot \frac{-1}{t^2}+\frac{-1}{\cos^2(3t+2f^2-v)}\cdot \frac{1}{2\sqrt{t}}=$$

$$=\frac{3}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}+\frac{4\frac{1}{t}}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}\cdot \frac{-1}{t^2}+\frac{-1}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}\cdot \frac{1}{2\sqrt{t}}=$$

$$=\frac{1}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}(3+\frac{4}{t}\cdot \frac{-1}{t^2}-\frac{1}{2\sqrt{t}})=$$

$$=\frac{1}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}(3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}})=$$

$$=\frac{3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}}}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}$$

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

Share with Friends