# Derivate of Function

I have to find the derivative of $\displaystyle g(z) = 1 + \sqrt{4-z}$.

I wrote $\displaystyle g(z)=1+(4-z)^{\frac{1}{2}}$.

Deriving - $\displaystyle g'(z)=\frac{1}{2}(4 - z)^{\frac{1}{2}-1}$ which then simplifies to be $\displaystyle \frac{1}{2}(4 - z)^{-\frac{1}{2}}$.

And finally $\displaystyle g'(z)=\frac{1}{2\sqrt{4-z}}$.

But the answer is positive instead of negative. What did I forget to do?

• What you forgot is The Chain Rule. – Gerry Myerson Feb 16 '16 at 12:03
• Two things - you can't simply delete constants, even though they vanish in derivative. Second, when you derive you have to multiply by inner derivative. In this case $\displaystyle y=\sqrt{f(x)} \Rightarrow y'=\frac{f'(x)}{2\sqrt{f(x)}}$. – Galc127 Feb 16 '16 at 12:08

## 2 Answers

Derivative of a function like $f(g(z))$ can be taken following the chain rule as; $$\frac{d(f(g(z)))}{dz} = \frac{\partial f(g(z))}{\partial g(z)} \frac{\partial g(z)}{\partial z}$$ where $f(g(z))=\sqrt{4-z}$ and $g(z)=4-z$ for your case.

You can find the source of negativity by following the chain rule as the second term in above equation.

You forgot the Chain Rule. You need to derive 4-z.