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How would I do these problems using Implicit Differentiation? I don't understand how to do them. Help?

Find $dy/dx$ if $\cos(4x)-2xe^{4y}=0$.


Find slope of the tangent line to the curve $\sqrt{12x+8y}+\sqrt{2xy}=12$ at the point $(4,2)$.

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You received some nice answers to your last question. You should accept one. –  user17794 Oct 1 '12 at 0:05
for the second one, it is -1. Just calculate the derivative and plug in values. –  user77399 May 11 '13 at 23:51
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2 Answers

Differentiate both sides of the equation with respect to $x$, remembering that $y$ is supposed to be a function of $x$. So, for example, the derivative of $y^2$ would be $2 y \dfrac{dy}{dx}$. Then solve for $\dfrac{dy}{dx}$. If you are asked for the slope at a point, substitute the $x$ and $y$ values for that point in your expression for $\dfrac{dy}{dx}$.

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I'm guessing this is homework, so I won't try to provide a complete solution to either of your questions.

If you differentiate your first equation with respect to $x$, you get $$ -4 \sin 4x - 2e^{4y} -8xye^{4y}\frac{dy}{dx} = 0 $$ which you should be able to rearrange.

If you differentiate the equation in your second question with respect to $x$, you get $$ \frac{12+8\frac{dy}{dx}}{2\sqrt{12x+8y}} + \frac{2y+2x\frac{dy}{dx}}{2\sqrt{2xy}} = 0 $$ and you can then substitute in your $x$ and $y$ values.

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Where did the -8xye^4y dy/dx come from? –  Taylor Oct 1 '12 at 0:26
You need the product rule and the chain rule. Be sure to treat $y$ as a function of $x$, not as a constant. –  user22805 Oct 1 '12 at 0:29
I'm still lost... –  Taylor Oct 1 '12 at 0:51
Well, if I'm going to differentiate $e^{4y}$ with respect to $x$, I need the chain rule. I'll multiply its derivative with respect to $y$ by the derivative of $y$ with respect to $x$. –  user22805 Oct 1 '12 at 1:03
Excuse me @DavidWallace, just a question. I could also apply the Implicit function theorem? –  Mark Oct 1 '12 at 16:46
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