Answer to Question #275145 in Calculus for Amu Raj

Question

2. a) Find the derivatives of the following functions with respect to x.

X^3+Y^3=3

Y= (sin x) ^tan x

Answer 1

Q1)

x³ + y³ = 3

The function is in implicit form.

Differentiating with respect to x

3x² + 3y² $\frac{dy}{dx}=0$

=> 3y² $\frac{dy}{dx}$ = -3x²

=> y² $\frac{dy}{dx}=$ -x²

=> $\frac{dy}{dx}= -\frac{x²}{y²}$

Q2

y = (sin x)^{tan x}

Taking logarithm of both sides

log_ey = log_e(sin x)^{tan x}

=> log_ey =tan x log_e(sin x)

Differentiating with respect to x

$\frac{1}{y}\frac{dy}{dx}= tanx\frac{d}{dx}log_{e}sinx+log_{e}sinx\frac{d}{dx}tanx$

$\frac{1}{y}\frac{dy}{dx}= tanx\frac{1}{sinx}cosx +log_{e}sinx.(sec²x)$

=> $\frac{1}{y}\frac{dy}{dx}= tanx.cotx +log_{e}sinx.(sec²x)$

=> $\frac{dy}{dx}$ = y (1 +sec²x log_esinx)

=> $\frac{dy}{dx}$ = (sin x)^{tan x}(1 + sec²x log_esinx)