Question

Use the Second Fundamental Theorem of Calculus, if needed, to calculate each of the derivatives given below.

$\frac{d^2}{d{x}^2}{\int }_{x^2}^1 \ln \left|t\right|dt$

Short answer

Ans: $\frac{d^2}{d{x}^2}{\int }_{x^2}^1 \ln \left|t\right|dt=-2\ln \left(\left|x^2\right|\right)-4$

Step-by-step solution

Step 1. Given information.

given, $\frac{d^2}{d{x}^2}{\int }_{x^2}^1 \ln \left|t\right|dt$

Step 2. The objective is to calculate the derivative. 

Now, if $f$is continuous on role="math" localid="1648715451199" $[a,b]$then for allrole="math" localid="1648715456581" $x\in [a,b]$.

role="math" $\frac{d}{dx}{\int }_a^{u\left(x\right)} f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)$

The derivative can be written as,

role="math" localid="1648715535214" $\begin{array}{r}\frac{d^2}{d{x}^2}{\int }_{x^2}^1 \ln \left|t\right|dt\\ =\frac{d^2}{d{x}^2}-{\int }_1^{x^2} \ln \left|t\right|dt\end{array}$

So,

$\begin{array}{r}f\left(u\right(t\left)\right)=\ln \left|t\right|\\ f\left(u\right(x\left)\right)=\ln \left(\left|x^2\right|\right)\\ u^{\mathrm{\prime }}\left(x\right)=2x\\ f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=2x\ln \left(\left|x^2\right|\right)\end{array}$

Step 3. The derivative expression can be written as, 

$\begin{array}{r}\frac{d^2}{d{x}^2}{\int }_{x^2}^1 \ln \left|t\right|dt\\ =\frac{d^2}{d{x}^2}-{\int }_1^{x^2} \ln \left|t\right|dt\\ =\frac{d}{dx}\left(\frac{d}{dx}-{\int }_1^{x^2} \ln \left|t\right|dt\right)\\ =\frac{d}{dx}\left(-2x\ln \left(\left|x^2\right|\right)\right)\\ =-2\frac{d}{dx}\left(x\ln \left(\left|x^2\right|\right)\right)\\ =-2\left(\ln \left(\left|x^2\right|\right)+x\left(2x\right)\frac{1}{x^2}\right)\\ =-2\left(\ln \left(\left|x^2\right|\right)+2x^2\frac{1}{x^2}\right)\\ =-2\ln \left(\left|x^2\right|\right)-2\left(2\right)\\ =-2\ln \left(\left|x^2\right|\right)-4\end{array}$

Therefore, the value islocalid="1648715687817" $-2\ln \left(\left|x^2\right|\right)-4$