Applying the Chain Rule for ($\sin(x) + 1$)

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?

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Commented Jun 24 at 11:06
• If $f(u)=u+1$ then $f'(u)=1$ hence $f'(\sin{x})=1$. Commented Jun 24 at 11:08
• Michael pointed out the problem. The mistake here is that, when you're forming the first part of the product ($f'(g(x))$), you're not differentiating something that has $\sin x$ but rather simply the function $f(u)=u+1$, whose derivative is just $1$. Commented Jun 24 at 11:12

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

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