Question

Find the derivative of the function.

\(g(x) = \int_1^{\cos x} {\sqrt(3){{1 - {t^2}}}dt} \)

Short answer

The value of the derivative for the integral\(g(x) = \int_1^{\cos x} {\sqrt(3){{1 - {t^2}}}dt} \)is\( - \sin {(x)^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}\).

Step-by-step solution

Function definition

In mathematics, a function is an expression, rule, or law that describes the connection between one variable and another variable. In mathematics, functions are everywhere, and they're crucial for articulating physical links in the sciences.

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus is a theorem that connects the concepts of differentiating (calculating the gradient) and integrating (calculating the slope) (calculating the area under the curve).

Let\(f(x) = \int_1^x {\sqrt(3){{1 - {t^2}}}dt} \).

Using this Fundamental theorem of calculus –

\({f^'}(x) = \sqrt(3){{1 - {x^2}}}\)

Applying the Chain rule

Define\(h\)as\(h(x) = \cos (x)\), so\({h^'}(x) = - \sin (x)\).

Now, it is seen that\(g(x) = f(h(x))\).

Using the chain rule –

\(\begin{array}{c}{g^'}(x) = {h^'}(x) \cdot {f^'}(h(x))\\ = - \sin (x) \cdot \sqrt(3){{1 - \cos {{(x)}^2}}}\\ = - \sin (x) \cdot \sqrt(3){{\sin {{(x)}^2}}}\\ = - \sin {(x)^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}\end{array}\)

Therefore, the derivative is obtained as\( - \sin {(x)^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}\).