Find $\displaystyle\frac{dy}{dx}$ given that$$\sqrt{3x^7+y^2}=\sin^2y+100xy.$$
Should I start off by squaring both sides to get rid of the radical on the left? And then start the derivative process? Thank you.
This is what I have so far:
Differentiating both sides with respect to $x$: $$\begin{align}\frac{21x^6+2y\displaystyle\frac{dy}{dx}}{2(3x^7+y^2)^{1/2}} &= 2\sin(y) \cos(y) \cdot \displaystyle\frac{dy}{dx} +100\left(y+x\displaystyle\frac{dy}{dx}\right)\end{align}$$ I think I had made the left side $1/2$ to get rid of the root and forgot to apply it to the right side of the equation. –
I'm at $$3x^7+y^2=(\sin^2y+100xy)^2$$ then I think I would start from left to right until they are all in their derivative form?
If we square both sides, then we could continue like this.
$$3x^7+y^2=(\sin^2y+100xy)^2.$$ Let's first differentiate the left hand side with respect to $x$. We get $$21x^6+2y\frac{dy}{dx}.$$ Now let's concentrate on the harder right hand side of the equation. If we were facing something like $[f(x)]^n$ and we differentiated with respect to $x$, that'd result in $n[f(x)]^{n-1}f'(x)$ by the chain rule. Our case is similar: $$\begin{align}\frac{d}{dx}(\sin^2y+100xy)^2=2(\sin^2y+100xy)\times \frac{d}{dx}(\sin^2y+100xy)\end{align}.$$ Can you finish it from there?
I hope that helps. If you have any questions please don't hesitate to ask.