fbpx
calculus online - Free exercises and solutions to help you succeed!

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)

Final Answer

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

Leave a Reply