Answer to Question #276505 in Calculus for RIYAD

Question

1.    Find the derivatives of the following functions with respect to x.

𝑥³ + 𝑦³ = 3   

𝑦 = (sin 𝑥)^{𝑡𝑎𝑛𝑥}

Answer 1

$𝑥^3 + 𝑦^3 = 3$

$y=(3-x^3)^{1/3}$

$3x^2+3y^2y'=0$

$y'=-\frac{x^2}{y^2}=-\frac{x^2}{(3-x^3)^{2/3}}$

$𝑦 = (sin 𝑥)^{𝑡𝑎𝑛𝑥}$

$lny=tanxln(sinx)$

$(lny)'=\frac{y'}{y}=\frac{ln(sinx)}{cos^2x}+\frac{tanxcosx}{sinx}=\frac{ln(sinx)}{cos^2x}+1$

$y'=(\frac{ln(sinx)}{cos^2x}+1)y=(\frac{ln(sinx)}{cos^2x}+1)(sin 𝑥)^{𝑡𝑎𝑛𝑥}$