Question

find the derivative of y with respect to x.
y = (x^6 ln x)^3
\frac{dy}{dx}=□

Explanation:

Step1: Apply chain - rule

Let $u = x^{6}\ln x$, then $y = u^{3}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y = u^{3}$, $\frac{dy}{du}=3u^{2}=3(x^{6}\ln x)^{2}$.

Step2: Apply product - rule to find $\frac{du}{dx}$

Since $u = x^{6}\ln x$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{6}$ and $v=\ln x$. We know that $u^\prime = 6x^{5}$ and $v^\prime=\frac{1}{x}$. So $\frac{du}{dx}=6x^{5}\ln x + x^{6}\cdot\frac{1}{x}=6x^{5}\ln x+x^{5}$.

Step3: Calculate $\frac{dy}{dx}$

Using $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, we substitute $\frac{dy}{du}$ and $\frac{du}{dx}$:
$\frac{dy}{dx}=3(x^{6}\ln x)^{2}(6x^{5}\ln x + x^{5})=3x^{12}\ln^{2}x(6x^{5}\ln x + x^{5})=3x^{17}\ln^{2}x(6\ln x + 1)$.

Answer:

$3x^{17}\ln^{2}x(6\ln x + 1)$