Question

Find $\frac{dy}{dx}$ when 𝑥 and 𝑦 are connected by the relation $sec(x+y)=xy$

Answer 1

Given : $sec(x+y)=xy$
Differentiating both sides w.r.t. 𝑥
We get,
$\frac{d}{dx}\left(\sec \right(x+y\left)\right)=\frac{d}{dx}\left(xy\right)$
$\Rightarrow \sec (x+y).\tan (x+y).\frac{d(x+y)}{dx}=y+x\frac{dy}{dx}$
$[\therefore \frac{d}{dx}(\sec x)=\sec x.\tan x,\frac{d\left(uv\right)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}]$
$\Rightarrow \sec (x+y).\tan (x+y),(1+\frac{dy}{dx})=y+x\frac{dy}{dx}$
$\Rightarrow \sec (x+y).\tan (x+y).\frac{d(x+y)}{dx}=y+x\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}\left[\sec \right(x+y).\tan (x+y)-x]$
$=y-\sec (x+y).\tan (x+y)$
$\Rightarrow \frac{dy}{dx}=\frac{y-\sec (x+y).\tan (x+y)}{\sec (x+y).\tan (x+y)-x}$
Hence, the value of $\frac{dy}{dx}$ is
$\frac{y-\sec (x+y).\tan (x+y)}{\sec (x+y).\tan (x+y)-x}$