Additional differential calculus problems

First, rewrite $f(x)$ as $$f(x) = (7x^5 + 2)^{-1/2}.$$ Next, apply the Power rule and Chain rule: \begin{eqnarray} \dfrac{df(x)}{dx} &=& (-1/2)(7x^5+2)^{-1/2 -1} \dfrac{d(7x^5+2)}{dx} \\ \ &=& (-1/2)(7x^5+2)^{-3/2} (35x^4 + 0) \\ \ &=& \dfrac{-35x^4}{2(7x^5+2)^{3/2}} \end{eqnarray}

First recall that $$\dfrac{d(\sin x)}{dx} = \cos x.$$ Next, apply the Power rule and Chain rule: \begin{eqnarray} \dfrac{df(x)}{dx} &=& 5(x + \sin x)^4 \dfrac{d(x + \sin x)}{dx} \\ \ &=& 5(x + \sin x)^4 (1 + \cos x). \end{eqnarray}

First, rewrite $f(x)$ as $$f(x) = \left((x -3)(x^2 + 4)^{-1}\right)^3.$$ Then apply the Power rule, Chain rule, and Product rule: \begin{eqnarray} f'(x) &=& 3 \left((x -3)(x^2 + 4)^{-1}\right)^2 \left((x -3)(x^2 + 4)^{-1}\right)'. \end{eqnarray} Since \begin{eqnarray} \dfrac{d}{dx}\left(\dfrac{x -3}{x^2 + 4} \right) &=& \dfrac{d}{dx}\left((x -3)(x^2 + 4)^{-1}\right)\\ \ &=& \dfrac{d(x-3)}{dx}(x^2 + 4)^{-1} + (x-3)\dfrac{d\left((x^2 + 4)^{-1}\right)}{dx} \\ \ &=& (1)(x^2 + 4)^{-1} + (x-3)(x^2+4)^{-2}\dfrac{d(x^2 + 4)}{dx} \\ \ &=& (x^2 + 4)^{-1} + \dfrac{x-3}{(x^2+4)^2} (2x + 0) \\ \ &=&\dfrac{1}{x^2 + 4} + \dfrac{2x^2-6x}{(x^2+4)^2} \\ \ &=& \dfrac{x^2+4 + 3x^2 - 6x}{(x^2+4)^2} \\ \ &=& \dfrac{4x^2 - 6x + 4}{(x^2+4)^2}, \end{eqnarray} we plug this result back into $f'(x)$ to get \begin{eqnarray} f'(x) &=& 3 \left(\dfrac{x-3}{x^2+4}\right)^2\left(\dfrac{x -3}{x^2 + 4} \right)' \\ \ &=& 3 \left(\dfrac{x-3}{x^2+4}\right)^2\dfrac{4x^2 - 6x + 4}{(x^2+4)^2}. \end{eqnarray}

\begin{eqnarray} f'(x) &=& (3x^4 - 7)' (2x + \cos x) + (3x^4 - 7) (2x + \cos x)' \\ \ &=& (12x^3 - 0) (2x + \cos x) + (3x^4 - 7) (2 - \sin x) \\ \ &=& 12x^3 (2x + \cos x) + (3x^4 - 7) (2 - \sin x). \end{eqnarray}

We repeatedly apply the Chain rule: \begin{eqnarray} f'(x) &=& ( \cos (\sin (\exp x)))' \\ \ &=& - \sin (\sin (\exp x))\, (\sin (\exp x))' \\ \ &=& - \sin (\sin (\exp x))\, (\cos (\exp x))\, (\exp x)' \\ \ &=& - \sin (\sin (\exp x))\, (\cos (\exp x))\, (\exp x). \end{eqnarray}

Solution 1 (use the chain rule): \begin{eqnarray} f'(x) &=& (\exp (3 \ln x))' \\ \ &=& (\exp (3 \ln x)) (3 \ln x)' \\ \ &=& (\exp (3 \ln x)) \dfrac{3}{x} \\ \ &=& (\exp (\ln x^3)) \dfrac{3}{x} \\ \ &=& x^3 \dfrac{3}{x} \\ \ &=& 3x^2. \end{eqnarray} Hmm. . . The answer looks surprisingly simply. This often suggests that there was an easier way to calculate the answer. Turns out, the trick is to simplify the function before taking the derivative.


Solution 2 (first, simplify the function):
Notice that $$f (x) = \exp (3 \ln x) = \exp (\ln x^3) = x^3.$$ A simple application of the Power rule immediately gives us $$f'(x) = 3x^2,$$ the same exact answer as what we got using the Chain rule.


Moral of the story: Simplify the function before taking the derivative. It often makes it easier to differentiate.