So, according to the chain rule, $$ \frac{d(f(g(x)))}{dx} = f'(g(x)) \cdot g'(x). $$ Now, if we considered $f(x) = x+1$ and $g(x) = \sin(x)$ then: $$ f(g(x)) = \sin(x)+1 $$

In this case, shouldn't the derivative be: $$ (\sin(x)+1)' * \sin'(x) = \cos(x) \cdot \cos(x) $$

Because $f'(g(x)) = (\sin(x)+1)'$ and $g'(x) = \sin'(x)$. Even if the function is not necessarily composite, doesn't the logic apply?


2 Answers 2


For the functions $f(x)=x+1$ and $g(x)=\sin(x)$ you find the derivatives

$$f'(x)=1,\quad g'(x)=\cos(x)$$

Note, that $f'(x)=1$ for any $x\in\mathbb{R}$. Therefore

$$f'(g(x))\cdot g'(x)=1\cdot\cos(x)=\cos(x)$$

since $g(x)=\sin(x)\in\mathbb{R}$.

Note that $$f'(g(x))\ne \left[f(g(x))\right]'$$

The left hand side is $f'$ evaluated at the point $g(x)$, the right hand side is the derivative of $f(g(x))$ evaluated at the point $x$. Especially

$$f'(g(x))\ne \left[f(g(x))\right]'=\left[\sin(x)+1\right]'$$


Indeed, setting $f(x)=x+1$ and $g(x)=\sin x$ yields $f(g(x)) = \sin x + 1$, so you correctly set your functions.

Now, for the actual derivative, you need the functions $g'(x)$ and $f'(x)$. You correctly calculated $g'(x)=\cos x$, but when calculating $f'(x)$, you need to forget that $g$ exists. $f'(x)=1$, independent of what $g$ is, or even if $g$ exists. Whenever $f(x)=x+1$, you have $f'(x)=1$.

So, if $f'(x)=1$ for all values of $x$, then $f'(g(x))$ must also be $1$, which means

$$f'(g(x))\cdot g'(x) = 1\cdot \cos x = \cos x.$$


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