Question

question given (y = f(u)) and (u = g(x)), find (\frac{dy}{dx}) by using leibnizs notation for the chain - rule: (\frac{dy}{dx}=\frac{dy}{du}cdot\frac{du}{dx}). (y = 5u^{4}-5), (u = 3sqrt{x}). provide your answer below: (\frac{dy}{dx}=square)

Explanation:

Step1: Differentiate y with respect to u

Using the power - rule $\frac{d}{du}(au^n)=nau^{n - 1}$, for $y = 5u^{4}-5$, we have $\frac{dy}{du}=20u^{3}$.

Step2: Differentiate u with respect to x

Given $u = 3\sqrt{x}=3x^{\frac{1}{2}}$. Using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$, we get $\frac{du}{dx}=\frac{3}{2}x^{-\frac{1}{2}}$.

Step3: Apply the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=20u^{3}$ and $\frac{du}{dx}=\frac{3}{2}x^{-\frac{1}{2}}$ into the chain - rule formula. Since $u = 3\sqrt{x}$, we have:
\[

$$\begin{align*} \frac{dy}{dx}&=(20u^{3})\cdot(\frac{3}{2}x^{-\frac{1}{2}})\\ &=(20(3\sqrt{x})^{3})\cdot(\frac{3}{2}x^{-\frac{1}{2}})\\ &=(20\times27x^{\frac{3}{2}})\cdot(\frac{3}{2}x^{-\frac{1}{2}})\\ &=(540x^{\frac{3}{2}})\cdot(\frac{3}{2}x^{-\frac{1}{2}})\\ &=810x^{\frac{3}{2}-\frac{1}{2}}\\ &=810x \end{align*}$$

\]

Answer:

$810x$