Calculate derivative of
$$f(x)=\log(2+sinx)$$

Question

Calculate derivative of 
$$f(x)=\log(2+sinx)$$

geerky42 · Answer

Perhaps it's best if you explain how you got that answer?

anonymous · Answer

yes that would be nice

geerky42 · Answer

@mr.singh

anonymous · Answer

@geerky can you explain it pls ?

geerky42 · Answer

You can "see" it as composition of functions: for example:

Let $g(x)$ be $\log(x)$ and $h(x)$ be $2+\sin(x)$

So $f(x)$ would be $g(h(x))$

So you can apply chain rule, where $f'(x) = g'(h(x))\cdot h'(x)$

You agree that  $g'(x) = \dfrac{1}{x}$ and $h'(x)=\cos x$, right?

So $f'(x) = g'(h(x))\cdot h'(x) = \left[\dfrac{1}{2+sinx}\right]\cdot\cos x\~~~~~~~~~=\boxed{\dfrac{\cos x}{2+\sin x}}$

geerky42 · Answer

Is that clear?

anonymous · Answer

hmm i need a little bit time to understand i think

geerky42 · Answer

You know chain rule, right? You basically derivative "outer," leave "inner" untouched, then you multiply it by derivative of "inner"

anonymous · Answer

yes i know it, ok  now its better thank you very much geerky

geerky42 · Answer

Ok welcome.