Question

given y = f(u) and u = g(x), find $\frac{dy}{dx}=f(g(x))g(x)$. y = csc u, u = $\frac{1}{x}+3x$ $\frac{dy}{dx}=square$

Explanation:

Step1: Find the derivative of $y = f(u)$ with respect to $u$

The derivative of $y=\csc u$ with respect to $u$ is $y^\prime=f^\prime(u)=-\csc u\cot u$.

Step2: Find the derivative of $u = g(x)$ with respect to $x$

Given $u=\frac{1}{x}+3x=x^{- 1}+3x$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $u^\prime=g^\prime(x)=-x^{-2}+3=-\frac{1}{x^{2}}+3$.

Step3: Apply the chain - rule $\frac{dy}{dx}=f^\prime(g(x))g^\prime(x)$

Substitute $u = g(x)=\frac{1}{x}+3x$ into $f^\prime(u)$ and multiply by $g^\prime(x)$.
$\frac{dy}{dx}=(-\csc(\frac{1}{x}+3x)\cot(\frac{1}{x}+3x))(-\frac{1}{x^{2}}+3)=(\frac{1}{x^{2}} - 3)\csc(\frac{1}{x}+3x)\cot(\frac{1}{x}+3x)$

Answer:

$(\frac{1}{x^{2}} - 3)\csc(\frac{1}{x}+3x)\cot(\frac{1}{x}+3x)$