Question

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

Answer 1

The proof is given below

Step-by-step explanation:

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.

Know More:

Q: If e^y (x+1) =1 then show that dy/dx = -e^y.

Click Here: https://brainly.in/question/3102659

Q: Siny=xsin(a+y) then prove dy/dx= sina/(1-2xcosa+a^2)

Click Here: https://brainly.in/question/5186653

Answer 2

dy/dx =   y/x proved

Step-by-step explanation:

Given: Sec (x^5 + y^5 /x^5 - y^5 ) = a

So Cos (x^5 - y^5 /x^5 + y^5 ) = a

(x^5 - y^5 /x^5 + y^5 ) = Cos⁻¹ a

x^5 - y^5 = Cos⁻¹ a (x^5 + y^5)

(1-Cos⁻¹ a) x^5 = (1+Cos⁻¹ a) y^5 -----------(1)

 x^5 / y^5 = (1+Cos⁻¹ a) / (1-Cos⁻¹ a) ------(2)

Differentiating, we get:

(1+Cos⁻¹ a) 5y^4 dy/dx =  (1-Cos⁻¹ a) 5x^4

  dy/dx =   (1-Cos⁻¹ a) 5x^4 / (1+Cos⁻¹ a) 5y^4

  dy/dx =   (1-Cos⁻¹ a) x^4 / (1+Cos⁻¹ a) y^4

  dy/dx =   (1-Cos⁻¹ a) x^5 / (1+Cos⁻¹ a) y^5 * y/x

    dy/dx =   (1-Cos⁻¹ a) x^5 / (1+Cos⁻¹ a) y^5 * y/x

Substituting from equation 2, we get:

      dy/dx =   (1-Cos⁻¹ a) (1+Cos⁻¹ a)/ (1-Cos⁻¹ a)(1+Cos⁻¹ a) y^5 * y/x

Therefore      dy/dx =   y/x. Hence proved.