Question

If ${\displaystyle {\log }_5\left(\frac{x^4+y^4}{x^4-y^4}\right)=2,\text{show that}\frac{\text{dy}}{\text{dx}}=\frac{12x^3}{13y^3}}$.

Answer 1

${\displaystyle {\log }_5\left(\frac{x^4+y^4}{x^4-y^4}\right)}$ = 2

${\displaystyle \frac{x^4+y^4}{x^4-y^4}}$ = 5²

${\displaystyle \frac{x^4+y^4}{x^4-y^4}}$ = 25

${\displaystyle \frac{x^4+y^4}{x^4-y^4}}$ = 25

x⁴ + y⁴ = 25(x⁴ – y)⁴ 

x⁴ + y⁴ = 25x⁴ – 25y⁴

∴ y⁴ + 25y⁴ = 25x⁴ − x⁴ 

26y⁴ = 24x⁴ 

Differentiating both sides w.r.t.x, we get

${\displaystyle 26\frac{d}{dx}{y}^4=24\frac{\text{d}}{\text{dx}}{x}^4}$

${\displaystyle 26.4y^3\frac{dy}{dx}=24.4x^3\frac{d}{dx}x}$

${\displaystyle 26y^3 \frac{dy}{dx}=24x^3}$

${\displaystyle \frac{dy}{dx}=\frac{24x^3}{26y^3}}$

${\displaystyle \frac{dy}{dx}=\frac{12x^3}{13y^3}}$