Question

d) \\( \frac { d } { d x } \left( \int _ { \tan x } ^ { x ^ { 2 } } \ln \left( u ^ { 2 } + 5 \
ight) d u \
ight) = \\)

Explanation:

Step1: Apply the Fundamental Theorem of Calculus and the Chain Rule

According to the Fundamental Theorem of Calculus, if \(F(x)=\int_{a}^{x}f(u)du\), then \(F^\prime(x) = f(x)\). For \(\int_{g(x)}^{h(x)}f(u)du=\int_{a}^{h(x)}f(u)du-\int_{a}^{g(x)}f(u)du\), its derivative is \(f(h(x))\cdot h^\prime(x)-f(g(x))\cdot g^\prime(x)\) by the Chain Rule. Here \(f(u)=\ln(u^{2}+5)\), \(h(x) = x^{2}\), and \(g(x)=\tan x\).

Step2: Calculate the derivatives of \(h(x)\) and \(g(x)\)

The derivative of \(h(x)=x^{2}\) is \(h^\prime(x) = 2x\) using the power rule \((x^{n})^\prime=nx^{n - 1}\). The derivative of \(g(x)=\tan x\) is \(g^\prime(x)=\sec^{2}x\) since \((\tan x)^\prime=\sec^{2}x\).

Step3: Substitute into the formula

Substitute \(h(x)\), \(h^\prime(x)\), \(g(x)\), and \(g^\prime(x)\) into \(f(h(x))\cdot h^\prime(x)-f(g(x))\cdot g^\prime(x)\). We get \(\ln((x^{2})^{2}+5)\cdot2x-\ln((\tan x)^{2}+5)\cdot\sec^{2}x\).

Answer:

\(2x\ln(x^{4}+5)-\sec^{2}x\ln(\tan^{2}x + 5)\)