Question

Help me make implicit differentiation to sin(x^2y^2)=x

Answer 1

Differentiate both sides with respect to x:\[\bf \cos(x^2y^2)*2xy^2+2x^2y \frac{ dy }{ dx }=1\]Now re-arrange and solve for \(\bf \frac{dy}{dx}\).

@Christos

Answer 2

There should be brackets around \(\bf 2xy^2+2x^2y \frac{dy}{dx}\). Sorry about that, I forgot to place them.

@Christos

Answer 3

$$\sin x^2y^2=x$$Differentiate both sides:$$\frac{d}{dx}\left(\sin x^2y^2\right)=\frac{d}{dx}x$$Simplify our left-hand side to \(1\) and use the chain-rule on our left:$$\cos x^2y^2\cdot\left(2xy^2+2x^2y\frac{dy}{dx}\right)=1$$All that's left to do is to isolate our \(dy/dx\); divide both sides by \(\cos x^2y^2\) to yield:$$2xy^2+2x^2y\frac{dy}{dx}=\frac1{\cos x^2y^2}$$Now subtract \(2xy^2\) to isolate our term of interest:$$2x^2y\frac{dy}{dx}=\frac1{\cos x^2y^2}-2xy^2$$Dividing by \(2x^2y\) gives us our final result:$$\frac{dy}{dx}=\frac1{2x^2y\cos x^2y^2}-\frac{y}x$$

Answer 4

Why my result is 1-cos(x^2y^2)2xy^2/(cos(x^2y^2)2yx^2) ?

Answer 5

dy/dx = (1-cos(x^2y^2)2xy^2)/(cos(x^2y^2)2yx^2)

Answer 6

They are equivalent. You distributed first, hence why it looks difference. Notice what happens if you distribute your division:$$\frac{1-2xy^2\cos(x^2y^2)}{2x^2y\cos(x^2y^2)}=\frac1{2x^2y\cos(x^2y^2)}-\frac{2xy^2\cos(x^2y^2)}{2x^2y\cos(x^2y^2)}$$Observe that our second term simplifies by cancelling \(2xy\cos(x^2y^2)\) in the numerator and denominator to yield:$$\frac1{2x^2y\cos(x^2y^2)}-\frac{2xy^2\cos(x^2y^2)}{2x^2y\cos(x^2y^2)}=\frac1{2x^2y\cos(x^2y^2)}-\frac{y}x$$i.e. my answer

Answer 7

different* not difference