Question about $\frac{\mathrm d}{\mathrm dx}(3\cos^3x)$ using the chain rule

Question

We are taking the derivative of $3\cos^3{x}$.

Using the chain rule I got $9\cos^{2}{x}\left(-3\sin{x}\right)$, but this was incorrect as the second multiplied term should be $-\sin x$.

My question is why is it $-\sin{x}$ instead of $-3\sin{x}$? According to the constant rule shouldn't it be $$3 \frac{\mathrm{d}}{\mathrm{d}x}\left(\cos x\right)?$$

Answer 1

Apply the constant rule first $(3\cos^3{x})' = 3(\cos^3{x})'$.

You have misapplied the chain rule by counting the constant "twice".

Answer 2

Let $f(x)=3\cos^3x$. Then $f(x)=(h\circ g)(x)$ where $g(x)=\cos x$ and $h(y)=3y^3$. The chain rule states that $$(h\circ g)'(x)=g'(x)(h'\circ g)(x).$$ Here $g'(x)=-\sin x$ and $h'(y)=9y^2$, so that $(h'\circ g)(x)=9\cos^2x$, etc.

Answer 3

Let $\cos x=u$ Hence $$\frac {d(3u^3)}{dx}= 9u^2\cdot \frac {du}{dx}$$ In our case $$\frac {du}{dx}=-\sin x$$ Hence $$\frac {d(3\cos^3 x)}{dx}=-9\cos^2 x\sin x$$

Answer 4

You are differentiating $f\circ g(x)$ where $f(x) = 3x^3$ and $g(x)=\cos x.$ By the chain rule, $(f\circ g)'(x) = f'(g(x))g'(x) = 9\cos^2x(-\sin x).$

I'm not sure where you're getting the extra $3$ from.