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
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})}
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