Question

Please open the image and help me

Answer 1

1.

Looks like

$y=2\sqrt[3]{x}+\dfrac1{x^2}+\pi$

Write $\sqrt[3]{x}$ as a fractional power, $x^{1/3}$. This makes it more obvious that the power rule should be used here.

$y'=(2x^{1/3})'+(x^{-2})'+\pi'$

$y'=\dfrac23x^{-2/3}-2x^{-3}$

$y'=\dfrac2{3\sqrt[3]{x^2}}-\dfrac2{x^3}$

2.

$y=(\sin(2x)+\tan(3x))^e$

Power and chain rule:

$y'=\left((\sin(2x)+\tan(3x))^e\right)'$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(\sin(2x)+\tan(3x))'$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(\cos(2x)(2x)'+\sec^2(3x)(3x)')$

$y'=e(\sin(2x)+\tan(3x))^{e-1}(2\cos(2x)+3\sec^2(3x))$