Aspire Faculty ID #13521 · Topic: JAMIA MILLIA ISLAMIA MCA 2024 · Just now
JAMIA MILLIA ISLAMIA MCA 2024

If $\sec!\left(\dfrac{x-y}{x+y}\right)=a$, then $\dfrac{dy}{dx}$ is

Solution

Let $u=\dfrac{x-y}{x+y}$. Since $\sec u=a$ (constant), $u' = 0$. $\displaystyle 0=\frac{d}{dx}!\left(\frac{x-y}{x+y}\right)=\frac{(1-y')(x+y)-(x-y)(1+y')}{(x+y)^2}$ $\Rightarrow 2y-2xy'=0\Rightarrow y'=\dfrac{y}{x}.$
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