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  • Find dy/dx when x and y are connected by the relation given sec (x + y) = xy

Find \frac{dy}{dx} when x and y are connected by the relation given

sec (x + y) = xy

Answers (1)

We have,

sec(x + y) = xy

By the rules given below:

Chain Rule

f(x) = g(h(x))

f’(x) = g’(h(x))h’(x)

Product rule:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}[\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})\\ &\text { On differentiating both sides with respect to } x, \text { we get }\\ &\sec (x+y) \tan (x+y) \frac{d}{d x}(x+y)=y+x \frac{d}{d x} y\\ &\left[\because \frac{d}{d x} \sec (x)=\sec x \tan x\right]\\ &\Rightarrow \sec (\mathrm{x}+\mathrm{y}) \tan (\mathrm{x}+\mathrm{y})\left[1+\frac{\mathrm{d} y}{\mathrm{~d} \mathrm{x}}\right]=\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} \end{aligned}

\\ \Rightarrow \sec (x+y) \tan (x+y)+\sec (x+y) \tan (x+y) \frac{d y}{d x}=y+x \frac{d y}{d x} \\ \Rightarrow \sec (x+y) \tan (x+y) \frac{d y}{d x}-x \frac{d y}{d x}=y-\sec (x+y) \tan (x+y) \\ \Rightarrow \frac{d y}{d x}[\sec (x+y) \tan (x+y)-x]=y-\sec (x+y) \tan (x+y) \\ \therefore \frac{d y}{d x}=\frac{y-\sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y)-x}

 

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