Question

find dy/dx given y = definite integral 1/(1+t^2) . upper cosx. lower 0. show steps/explain.

Answer 1

Hello,

If i have well understood:

$F(x)= \int\limits^{b(x)}_{a(x)} {f(x,t)} \, dt \\ \boxed{ F'(x)=f(x,b(x))*b'(x)-f(x,a(x))*a'(x)+ \int\limits^{b(x)}_{a(x)} { \frac{\delta f(x,t)}{\delta x}} \, dt}\\\\ F(x)= \int\limits^{cos(x)}_{0} { \dfrac{1}{1+t^2} } \, dt \\ F'(x)= \dfrac{1}{1+cos^2(x)}*(-sin(x))- \frac{1}{1}*0+ \int\limits^{cos(x)}_{0} { \dfrac{\delta \dfrac{1}{1+t^2}}{\delta x} \, dt \\ =-\ \dfrac{sin(x)}{1+cos^2(x)}$

Answer 2

Answer:

$\displaystyle y' = \frac{-\sin x}{1 + \cos^2 x}$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Rule [Chain Rule]:                                                                                 $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integration

• Integrals

Integration Rule [Fundamental Theorem of Calculus 2]:                                     $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \int\limits^{\cos x}_0 {\frac{1}{1 + t^2}} \, dt$

Step 2: Differentiate

1. Chain Rule:                                                                                                     $\displaystyle y' = \frac{d}{dx} \bigg[ \int\limits^{\cos x}_0 {\frac{1}{1 + t^2}} \, dt \bigg] \cdot \frac{d}{dx}[\cos x]$
2. Fundamental Theorem of Calculus 2:                                                         $\displaystyle y' = \frac{1}{1 + \cos^2 x} \cdot \frac{d}{dx}[\cos x]$
3. Trigonometric Differentiation:                                                                       $\displaystyle y' = \frac{1}{1 + \cos^2 x} \cdot -\sin x$
4. Simplify:                                                                                                         $\displaystyle y' = \frac{-\sin x}{1 + \cos^2 x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration