Question

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

$\frac{d}{dx}{\int }_1^x \ln tdt$

Short answer

Ans: $\frac{d}{dx}{\int }_1^x \ln tdt=\ln x$

Step-by-step solution

Step 1. Given information.

given expression, $\frac{d}{dx}{\int }_1^x \ln tdt$

Step 2. The objective is to calculate the derivative. 

Now, if $f$is continuous on $[a,b]$ then for all $x\in [a,b]$,

$\frac{d}{dx}{\int }_a^x f\left(t\right)dt=f\left(x\right)$

So,

$\begin{array}{r}f\left(t\right)=\ln t\\ f\left(x\right)=\ln x\end{array}$

Step 3. The derivative expression can be written as, 

$\begin{array}{r}\frac{d}{dx}{\int }_1^x \ln tdt\\ =\ln x\left[f\right(x)=\ln x\rfloor \end{array}$

Therefore, the value is $\ln x$.