Question

question 12
find $\frac{dy}{dx}$ where $y^{4}sin(7x)+2x^{7}-y^{8}=3$
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score: 40/60
answered: 4/15
homework 3.8 implicit differentiation

Explanation:

Step1: Differentiate each term

Differentiate $y^{4}\sin(7x)+2x^{7}-y^{8}=3$ term - by - term with respect to $x$.
For $y^{4}\sin(7x)$, use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = y^{4}$ and $v=\sin(7x)$. The derivative of $y^{4}$ with respect to $x$ is $4y^{3}\frac{dy}{dx}$ (by the chain - rule), and the derivative of $\sin(7x)$ with respect to $x$ is $7\cos(7x)$. So the derivative of $y^{4}\sin(7x)$ is $4y^{3}\sin(7x)\frac{dy}{dx}+7y^{4}\cos(7x)$. The derivative of $2x^{7}$ with respect to $x$ is $14x^{6}$, and the derivative of $-y^{8}$ with respect to $x$ is $-8y^{7}\frac{dy}{dx}$, and the derivative of the constant 3 is 0.
So we have $4y^{3}\sin(7x)\frac{dy}{dx}+7y^{4}\cos(7x)+14x^{6}-8y^{7}\frac{dy}{dx}=0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$4y^{3}\sin(7x)\frac{dy}{dx}-8y^{7}\frac{dy}{dx}=- 14x^{6}-7y^{4}\cos(7x)$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(4y^{3}\sin(7x)-8y^{7})=-14x^{6}-7y^{4}\cos(7x)$.
Then $\frac{dy}{dx}=\frac{-14x^{6}-7y^{4}\cos(7x)}{4y^{3}\sin(7x)-8y^{7}}=\frac{7y^{4}\cos(7x) + 14x^{6}}{8y^{7}-4y^{3}\sin(7x)}$.

Answer:

$\frac{7y^{4}\cos(7x)+14x^{6}}{8y^{7}-4y^{3}\sin(7x)}$