Question

1. use part 1 of the fundamental theorem of calculus to find the derivative of the function for

a. \\( g(x) = \int_{0}^{x^3} \cos(t^2) \\, dt \\)
b. \\( f(x) = \int_{x}^{0} \sqrt{1 + \sec t} \\, dt \\)
c. \\( h(x) = \int_{1}^{\sqrt{x}} \frac{z^2}{z^4 + 1} \\, dz \\)
d. \\( y = \int_{2x}^{x^2} e^t \\, dt \\)

Explanation:

Part (a)

Step1: Recall Fundamental Theorem of Calculus Part 1 and Chain Rule

The Fundamental Theorem of Calculus Part 1 states that if \( g(x)=\int_{a}^{u(x)} f(t) dt \), then \( g'(x) = f(u(x))\cdot u'(x) \). Here, \( a = 0 \), \( u(x)=x^{3} \), and \( f(t)=\cos(t^{2}) \).

Step2: Apply the Theorem and Chain Rule

First, find the derivative of the upper limit \( u(x)=x^{3} \), so \( u'(x) = 3x^{2} \). Then, substitute \( u(x) \) into \( f(t) \): \( f(u(x))=\cos((x^{3})^{2})=\cos(x^{6}) \). Now, multiply by \( u'(x) \): \( g'(x)=\cos(x^{6})\cdot 3x^{2}=3x^{2}\cos(x^{6}) \).

Part (b)

Step1: Rewrite the Integral Using Property of Integrals

Recall that \( \int_{x}^{0} f(t) dt=-\int_{0}^{x} f(t) dt \). So, \( F(x)=-\int_{0}^{x} \sqrt{1 + \sec t} dt \). Now, we can apply the Fundamental Theorem of Calculus Part 1.

Step2: Apply the Theorem

Let \( f(t)=\sqrt{1+\sec t} \) and \( u(x)=x \) (so \( u'(x) = 1 \)). Then, by the theorem, the derivative of \( -\int_{0}^{x} f(t) dt \) is \( -f(x)\cdot 1 \). Substitute \( x \) into \( f(t) \): \( f(x)=\sqrt{1+\sec x} \). So, \( F'(x)=-\sqrt{1 + \sec x} \).

Part (c)

Step1: Identify the Form and Apply the Theorem

The function is \( h(x)=\int_{1}^{\sqrt{x}} \frac{z^{2}}{z^{4}+1} dz \). Let \( u(x)=\sqrt{x}=x^{\frac{1}{2}} \), \( a = 1 \), and \( f(z)=\frac{z^{2}}{z^{4}+1} \). First, find \( u'(x) \): \( u'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \).

Step2: Substitute into the Formula

Substitute \( u(x) \) into \( f(z) \): \( f(u(x))=\frac{(\sqrt{x})^{2}}{(\sqrt{x})^{4}+1}=\frac{x}{x^{2}+1} \). Then, multiply by \( u'(x) \): \( h'(x)=\frac{x}{x^{2}+1}\cdot\frac{1}{2\sqrt{x}} \). Simplify: \( \frac{x}{x^{2}+1}\cdot\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}}{2(x^{2}+1)} \) (since \( \frac{x}{\sqrt{x}}=\sqrt{x} \)).

Part (d)

Answer:

s:
a. \( \boldsymbol{g'(x)=3x^{2}\cos(x^{6})} \)
b. \( \boldsymbol{F'(x)=-\sqrt{1 + \sec x}} \)
c. \( \boldsymbol{h'(x)=\frac{\sqrt{x}}{2(x^{2}+1)}} \)
d. \( \boldsymbol{y' = 2x e^{x^{2}}-2e^{2x}} \)