Question

1. $y = 5u^{3}+1$, $u = 4x^{3}+3e^{x^{2}}$, find $\frac{dy}{du}$, $\frac{du}{dx}$, $\frac{dy}{dx}$

Explanation:

Step1: Find $\frac{dy}{du}$

Since $y = 5u^{3}+1$, using the power - rule for differentiation $\frac{d}{du}(au^{n})=nau^{n - 1}$, we have $\frac{dy}{du}=15u^{2}$.

Step2: Find $\frac{du}{dx}$

Given $u = 4x^{3}+3e^{x^{2}}$. The derivative of $4x^{3}$ using the power - rule is $12x^{2}$, and for $3e^{x^{2}}$, using the chain - rule. Let $t=x^{2}$, then $\frac{d}{dx}(3e^{x^{2}})=3e^{x^{2}}\cdot2x$. So $\frac{du}{dx}=12x^{2}+6xe^{x^{2}}$.

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

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=15u^{2}$ and $u = 4x^{3}+3e^{x^{2}}$ into it, we get $\frac{dy}{dx}=15(4x^{3}+3e^{x^{2}})^{2}(12x^{2}+6xe^{x^{2}})$.

Answer:

$\frac{dy}{du}=15u^{2}$, $\frac{du}{dx}=12x^{2}+6xe^{x^{2}}$, $\frac{dy}{dx}=15(4x^{3}+3e^{x^{2}})^{2}(12x^{2}+6xe^{x^{2}})$