Question

if x=sec(x^5+y^5/x^5-y^5)=a^2
then show dy/dx=y/x​

Answer 1

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Step-by-step explanation:

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Answer 2

The proof is given below:

Given

$\sec(\frac{x^5+y^5}{x^5-y^5})=a$

$\implies \cos(\frac{x^5-y^5}{x^5+y^5})=a$

$\implies \frac{x^5-y^5}{x^5+y^5}=\cos^{-1}a$

$\implies x^5-y^5=\cos^{-1}a(x^5+y^5)$

$\implies (1-\cos^{-1}a)x^5=(1+\cos^{-1}a)y^5$

$\implies (1+\cos^{-1}a)y^5=(1-\cos^{-1}a)x^5$  .............. (1)

Differentiating both sides w.r.t. x

$(1+\cos^{-1}a)5y^4\frac{dy}{dx}=(1-\cos^{-1}a)5x^4$

$\implies \frac{dy}{dx}=\frac{(1-\cos^{-1}a)5x^4}{(1+\cos^{-1}a)5y^4}$

$\implies \frac{dy}{dx}=\frac{(1-\cos^{-1}a)x^4}{(1+\cos^{-1}a)y^4}$

$\implies \frac{dy}{dx}=\frac{(1-\cos^{-1}a)x^5}{(1+\cos^{-1}a)y^5}\times\frac{y}{x}$

$\implies \frac{dy}{dx}=\frac{(1-\cos^{-1}a)}{(1+\cos^{-1}a)}\times\frac{(1+\cos^{-1}a}{(1-\cos^{-1}a)}\times\frac{y}{x}$ (putting value of $x^5/y^5$from eq (1))

$\implies \frac{dy}{dx}=\frac{y}{x}$                     (Proved)

Hope this answer is helpful.

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