Finding $\frac{d}{dx}(\sqrt{3\cos^3x})$

Question

I am attempting to find the derivative of $\sqrt{3\cos^3 x}$.

Using the chain rule, I got:

$$\frac{1}{2} \sqrt{(3\cos x)^3} × 3(3\cos x)^2 × (-3\sin x).$$

Apparently this is where I went wrong, the correct answer stated that instead of the derivative of the inner function should have been $-\sin x$ instead of $-3\sin x$. To my understanding the derivative of $3\cos x$ is $-3\sin x$. Is my reasoning incorrect?

Answer 1

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{3\cos^3x} &= \sqrt{3}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos^\frac32x\\ &= \sqrt{3}\cdot\frac32\cos^{\frac32 - 1}x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\cos x\\ &= \sqrt{3}\cdot\frac32\cos^\frac12x\cdot(-\sin x) \\ &= -\frac{3^\frac32}{2}\sin x\sqrt{\cos x} \end{align*}$$

Answer 2

Here is a correct implementation of chain rule

$\frac{d\sqrt{3cos^3(x)}}{dx} = \frac{1}{2\sqrt{3cos^3(x)}}*\frac{d( {3cos^3(x)})}{dx}$ = $\frac{1}{2\sqrt{3cos^3(x)}}*3*3*cos^2(x)*\frac{d({cos(x)})}{dx}$

$\qquad \qquad = \frac{1}{2\sqrt{3cos^3(x)}}*9*cos^2(x)*(-sin(x)) = -\frac{3^{\frac{3}{2}}}{2}*sin(x)*\sqrt{cos(x)}$

Answer 3

You have to be more careful with negative exponents.

$$ d/dx (\sqrt{3\cos^3 x}) = d/dx (3\cos^3 x)^{1/2}= $$

$$ (1/2)(3\cos^3 x)^{-1/2}(9\cos^2 x )(-\sin x)=$$

$$(-3 \sqrt 3/2)(\cos ^{1/2}x)\sin x =$$

$$(-3 \sqrt 3/2)\sin x \sqrt {\cos x} $$

Answer 4

For this kind of problems, logarithmic differentiation makes life much easier $$y=\sqrt{3\cos^3 (x)}\implies \log(y)=\log(\sqrt{3})+\frac 32 \log(\cos(x))$$ Differentiate both sides $$\frac{y'} y=-\frac 3 2 \frac{\sin(x)}{\cos(x)}$$ $$y'=y \times \frac{y'} y= ???$$ Simplify.