derivative of $\frac{2}{3}x^{3-e}$

Question

Find the derivative:$\;\;\;\;\;\;\dfrac{2}{3}x^{3-e}$

I am not sure how to solve this problem. My try:

$\ln y=\dfrac{2}{3}(3-e)\ln x$
$\dfrac{1}{y}\times y\;'=\dfrac{2}{3}(3-e)\dfrac{1}{x}$

I'm not sure how to proceed from here. The answer is $\dfrac{2}{3}x^{2-e}(e-3)$ and I don't see how to arrive at that answer.

Answer 1

Let $f\colon\Bbb{R}\to\Bbb{R}$, with $f(x)=\frac{2}{3}x^{3-e}$, then $$ \frac{\mathrm{d}f}{\mathrm{d}x}= \frac{2}{3}\frac{\mathrm{d}}{\mathrm{d}x}\Big(x^{3-e}\Big) = \frac{2}{3}(3-e)x^{3-e-1} = \frac{2(3-e)}{3}x^{2-e} $$

Answer 2

we have $y=\frac{2}{3}x^{3-e}$ taking the logarithm of both sides we get $\ln(y)=\ln\left(\frac{2}{3}\right)+(3-e)\ln(x)$ differentiating with respect to $x$ we find $\frac{y'}{y}=\frac{3-e}{x}$ this is equivalent to $y'=\frac{3-e}{x}\frac{2}{3}x^{3-e}$