Differentiate $f(x)=\int_{\sqrt{x}}^{\arcsin x} \ln\theta d \theta$
Answer
You need to use the Leibniz rule:
\[(\int_{g(x)}^{h(x)}f(\theta)d\theta)'=h'(x)f(h(x))-g'(x)f(g(x)).\]
Using this formula we get
\[f'(x)=(\arcsin x)' \ln(\arcsin x)-(x^{1/2})'\ln(x^{1/2})\]
\[=\frac{1}{\sqrt{1-x^2}} \ln(\arcsin x)-\frac{1}{2}x^{-1/2}\ln (x^{1/2}).\]
\[(\int_{g(x)}^{h(x)}f(\theta)d\theta)'=h'(x)f(h(x))-g'(x)f(g(x)).\]
Using this formula we get
\[f'(x)=(\arcsin x)' \ln(\arcsin x)-(x^{1/2})'\ln(x^{1/2})\]
\[=\frac{1}{\sqrt{1-x^2}} \ln(\arcsin x)-\frac{1}{2}x^{-1/2}\ln (x^{1/2}).\]
Erdos
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