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$$x\sin(4x+5y)=y\cos(x)$$ I am trying to use implicit differentiation to find dx/dy for this problem but the answer i keep getting is $$4x\cos(4x+5y)=-y\sin(x)$$ and I am stuck.

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  • $\begingroup$ You need to use the product rule when you differentiate both sides. $\endgroup$ – user84413 Mar 8 '14 at 0:01
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You're forgetting that $x$ is dependent on $y$ so you have to take chain rule into account. I'll do part of it. Hopefully you can see what to do from there.

$$\begin{align}\frac{d}{dy}(x\sin(4x+5y)) &\stackrel{\text{product rule}}{=} \left(\frac{d}{dy}x\right)\sin(4x+5y)+x\frac{d}{dy}\sin(4x+5y)\\ &\stackrel{\text{chain rule}}{=} \frac{dx}{dy}\sin(4x+5y) + x\cos(4x+5y)\cdot\frac{d}{dy}(4x+5y)\\ &= \frac{dx}{dy}\sin(4x+5y)+x\cos(4x+5y)\left(4\frac{dx}{dy}+5\right)\end{align}$$

Does this make sense?

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  • $\begingroup$ is this just the left side of the equation $\endgroup$ – Kelli Davis Mar 8 '14 at 0:41
  • $\begingroup$ Yes it is. You should have a clearer idea of what to do with the right side, I hope. $\endgroup$ – Cameron Williams Mar 8 '14 at 0:43
  • $\begingroup$ i'm still having problems. can you break it down a little more? $\endgroup$ – Kelli Davis Mar 8 '14 at 2:46
  • $\begingroup$ What specifically are you having trouble with? $\endgroup$ – William Chang Mar 8 '14 at 5:06
  • $\begingroup$ @Mathster I used the product and chain rule on both sides. I keep getting (sin(4x+5y) + 4xcos(4x+5y)+ysinx)/cosx. but that isnt the right answer $\endgroup$ – Kelli Davis Mar 8 '14 at 5:29
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$$\frac{d}{dx}\left(x\sin(4x+5y)=y\cos(x)\right)\\ \implies \sin(4x+5y)+x\frac{d}{dx}\sin(4x+5y)=\cos(x)\frac{dy}{dx}+y\frac{d}{dx}\cos(x)$$ Let us differentiate part-by-part. $$\frac{d}{dx}\sin(4x+5y)=\frac{d}{d(4x+5y)}(\sin(4x+5y))\cdot\left(20y+20x\frac{dy}{dx}\right)=\\20\cos(4x+5y)\left(y+x\frac{dy}{dx}\right)$$ Then, $$\frac{d}{dx}\cos(x)=-\sin(x)$$ Substitution gives $$\sin(4x+5y)+x\frac{d}{dx}\sin(4x+5y)=\cos(x)\frac{dy}{dx}+y\frac{d}{dx}\cos(x)\\ \implies\sin(4x+5y)+20x\cos(4x+5y)\left(y+x\frac{dy}{dx}\right)=\cos(x)\frac{dy}{dx}-y\sin(x)\\ \implies \sin(4x+5y)+20x\cos(4x+5y)y+y\sin(x)=\left(\cos(x)-20x^2\cos(4x+5y)\right)\frac{dy}{dx}\\ \implies\boxed{\dfrac{dy}{dx}=\dfrac{\sin(4x+5y)+20x\cos(4x+5y)y+y\sin(x)}{\cos(x)-20x^2\cos(4x+5y)}}$$

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  • $\begingroup$ this is not the right answer $\endgroup$ – Kelli Davis Mar 8 '14 at 5:03

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