Differentiate $\tan(xy)= y+2$

Question

Here is what I did:

$$\tan(xy)=y+2$$

$$(xy')(y)\sec^2(xy)=y'$$

Now I'm stuck on simplifying this. How do I get all the y's on one side and divide?

Answer 1

There's an error in what you've done.

Taking the derivative of both sides yields $$\text{sec}^2(xy)\left( xy' +y \right) = y'$$

Now solve for $y'$.

$$\text{sec}^2(xy)xy' - y' = -y\cdot \text{sec}^2(xy)$$ $$\Rightarrow y'\left(x\cdot\text{sec}^2(xy)-1\right) = -y\cdot \text{sec}^2(xy)$$ $$\Rightarrow y' = \frac{-y\cdot \text{sec}^2(xy)}{x\cdot\text{sec}^2(xy)-1}$$.

Answer 2

Hint: $\sec^2z=\tan^2z+1$ - eliminate the trigonometric terms

Answer 3

$$\tan (xy)=y+2$$ $$\sec^2 (xy)\left(y+x\frac{dy}{dx}\right)=\frac{dy}{dx}$$

$$y\sec^2 (xy)=\left(1-x\sec^2(xy)\right)\frac{dy}{dx}$$ $$\frac{dy}{dx}=\frac{y\sec^2 (xy)}{1-x\sec^2(xy)}$$ and multiplying numerator and denominator by $\cos^2(xy)$ we get

$$\frac{dy}{dx}=\frac{y}{\cos^2(xy)-x}$$

Answer 4

A lazy way : express $$x=\frac{\tan ^{-1}(y+2)}{y}$$ So, $$\frac{dx}{dy}=\frac{1}{y \left((y+2)^2+1\right)}-\frac{\tan ^{-1}(y+2)}{y^2}$$ Now compute $\frac{dy}{dx}$