Question

If given $g\left(x\right)={\int }_2^{x^2}\frac{1}{ln(1+t^2)}dtthenfind{g}^{\prime }\left(\sqrt{2}\right)$ .

• A. $\frac{2.\sqrt{2}}{ln\left(2\right)}$
• B. $\frac{2.\sqrt{2}}{ln\left(3\right)}$
• C. $\frac{\sqrt{2}}{ln\left(3\right)}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{2.\sqrt{2}}{ln\left(3\right)}$

We can see that the function g(x) is given in terms of integral. So to differentiate it we’ll have to apply Leibnitz formula.
$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(t\right)dt=f\left(b\right(x\left)\right).b^{\prime }\left(x\right)-f\left(a\right(x\left)\right).a^{\prime }\left(x\right)So,g^{\prime }\left(x\right)=\frac{1}{ln(1+x^2)}.2x-\frac{1}{ln(1+2^2)}.0g^{\prime }\left(x\right)=\frac{2x}{ln(1+x^2)}And{g}^{\prime }\left(\sqrt{2}\right)=\frac{2.\sqrt{2}}{ln\left(3\right)}$