Question

$(3x^4-x^3+4{)}^{5/2}$ differentiate w.r.t x.

Answer 1

Let $y=(3x^4-x^3+4{)}^{\frac{5}{2}}$

Differentiate on both sides w.r.t x

$\frac{dy}{dx}=\frac{5}{2}(3x^4-x^3+4{)}^{\frac{3}{2}}\ast \frac{d}{dx}(3x^4-x^3+4)$

$=\frac{5}{2}(3x^4-x^3+4{)}^{\frac{3}{2}}\ast (12x^3-3x^2)$

$\therefore \frac{dy}{dx}=\frac{15x^2}{2}(4x-1)(3x^4-x^3+4{)}^{\frac{3}{2}}$