Question

lf $x^3={\int }_0^z\frac{1}{\sqrt{1+6t^2}}$ dt then $\frac{dz}{dx}$ when $x=1,z=2$

• A. $-15$
• B. $15$
• C. $\frac{1}{15}$
• D. $-\frac{1}{15}$

Answer 1

The correct option is B $15$

Leibniz rule
$3x^{2}dx=\frac{1}{\sqrt{1+6z^2}}dz$
$(3\times 1)\left(\sqrt{1+6\times 4}\right)=\frac{dz}{dx}{|}_{x=1,z=2}$

$15=\frac{dz}{dx}{|}_{x=1,z=2}$