What is the derivative of $tan (x y)$ $tan(xy)$ ?

Question

26088 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-30T08:56:57

$\frac{d}{d x} (tan (x y)) = {sec}^{2} (x y) y$ $frac{d}{dx}(tan (xy))=\sec ^2(xy)y$

$\frac{d}{d x} (tan (x y))$ $frac{d}{dx}(tan (xy))$

Applying Chain rule,
$\frac{d f (u)}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

Let $x y = u$ $xy=u$

$= \frac{d}{d u} (tan (u)) \frac{d}{d x} (x y)$ $=\frac{d}{du}(\tan (u))\frac{d}{dx}(xy)$

We know,
$\frac{d}{d u} (tan (u)) = {sec}^{2} (u)$ $\frac{d}{du}(\tan (u))=\sec ^2(u)$ and,
$\frac{d}{d x} (x y) = y$ $\frac{d}{dx}(xy)=y$

So,
$= {sec}^{2} (u) y$ $=\sec ^2(u)y$

Finally,substituting back, $x y = u$ $xy=u$
$= {sec}^{2} (x y) y$ $=\sec ^2(xy)y$

What is the derivative of tan(xy)tan(xy)?