Question

find $\frac{dy}{du}$, $\frac{du}{dx}$, and $\frac{dy}{dx}$. $y = u^{54}$ and $u = 3x^{5}+5x^{3}$ $\frac{dy}{du}=square$

Explanation:

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

Using the power - rule for differentiation $\frac{d}{du}(u^n)=nu^{n - 1}$, for $y = u^{54}$, we have $\frac{dy}{du}=54u^{53}$.

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

For $u = 3x^{5}+5x^{3}$, using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $\frac{du}{dx}=3\times5x^{4}+5\times3x^{2}=15x^{4}+15x^{2}$.

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

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=54u^{53}$ and $\frac{du}{dx}=15x^{4}+15x^{2}$ into the chain - rule formula. Since $u = 3x^{5}+5x^{3}$, we have $\frac{dy}{dx}=54(3x^{5}+5x^{3})^{53}(15x^{4}+15x^{2})$.

Answer:

$\frac{dy}{du}=54u^{53}$, $\frac{du}{dx}=15x^{4}+15x^{2}$, $\frac{dy}{dx}=54(3x^{5}+5x^{3})^{53}(15x^{4}+15x^{2})$