Question

1. - / 1 points

find the derivative of the function.
g(x) = \\( \int _ { \tan ( x ) } ^ { 3 x ^ { 2 } } \frac { 1 } { \sqrt { 7 + t ^ { 4 } } } d t \\)
g ^ { \prime } ( x ) =

Explanation:

Step1: Apply the fundamental theorem of calculus and chain rule

If \(g(x)=\int_{a(x)}^{b(x)}f(t)dt\), then \(g^{\prime}(x)=f(b(x))\cdot b^{\prime}(x)-f(a(x))\cdot a^{\prime}(x)\). Here \(f(t)=\frac{1}{\sqrt{7 + t^{4}}}\), \(b(x)=3x^{2}\), \(a(x)=\tan(x)\).

Step2: Calculate \(b^{\prime}(x)\) and \(a^{\prime}(x)\)

Differentiate \(b(x)=3x^{2}\), \(b^{\prime}(x) = 6x\). Differentiate \(a(x)=\tan(x)\), \(a^{\prime}(x)=\sec^{2}(x)\).

Step3: Substitute into the formula

\(g^{\prime}(x)=\frac{1}{\sqrt{7+(3x^{2})^{4}}}\cdot6x-\frac{1}{\sqrt{7+\tan^{4}(x)}}\cdot\sec^{2}(x)\)
Simplify \((3x^{2})^{4}=81x^{8}\), so \(g^{\prime}(x)=\frac{6x}{\sqrt{7 + 81x^{8}}}-\frac{\sec^{2}(x)}{\sqrt{7+\tan^{4}(x)}}\)

Answer:

\(g^{\prime}(x)=\frac{6x}{\sqrt{7 + 81x^{8}}}-\frac{\sec^{2}(x)}{\sqrt{7+\tan^{4}(x)}}\)