Question

this is similar to section 2.8 problem 14: given y = √u and u = 2x^7+2 :
\frac{dy}{du}=
\frac{du}{dx}=
\frac{dy}{dx}=
hint: follow example 2. the variable in the first answer is u.
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Explanation:

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

Differentiate $y = \sqrt{u}=u^{\frac{1}{2}}$ with respect to $u$. Using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

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

Differentiate $u = 2x^{7}+2$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$ and $\frac{d}{dx}(c)=0$ (where $c$ is a constant), we get $\frac{du}{dx}=14x^{6}$.

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

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$ and $\frac{du}{dx}=14x^{6}$ into the chain - rule formula. Since $u = 2x^{7}+2$, we have $\frac{dy}{dx}=\frac{1}{2\sqrt{2x^{7}+2}}\cdot14x^{6}=\frac{7x^{6}}{\sqrt{2x^{7}+2}}$.

Answer:

$\frac{dy}{du}=\frac{1}{2\sqrt{u}}$
$\frac{du}{dx}=14x^{6}$
$\frac{dy}{dx}=\frac{7x^{6}}{\sqrt{2x^{7}+2}}$