Question

Find the derivative of the function

Answer 1

Answer:

$\displaystyle y' = \frac{9 \bigg[ 6x^\big{\frac{9}{4}} \sqrt{x^3} \sin (x^3) - \sin (\sqrt{x}) \bigg] }{2x^\big{\frac{1}{4}}}$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]:                                                         $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

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]:                                     $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Integration Property [Multiplied Constant]:                                                         $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Flipping Integral]:                                                               $\displaystyle \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)} \, dx$

Integration Property [Splitting Integral]:                                                               $\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \int\limits^{x^3}_{\sqrt{x}} {9\sqrt{t} \sin (t)} \, dt$

Step 2: Differentiate

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 $\displaystyle y = 9\int\limits^{x^3}_{\sqrt{x}} {\sqrt{t} \sin (t)} \, dt$
2. [Integral] Rewrite [Integration Property - Splitting Integral]:                      $\displaystyle y = 9 \bigg[ \int\limits^0_{\sqrt{x}} {\sqrt{t} \sin (t)} \, dt + \int\limits^{x^3}_0 {\sqrt{t} \sin (t)} \, dt \bigg]$
3. [1st Integral] Rewrite [Integration Property - Flipping Integral]:                 $\displaystyle y = 9 \bigg[ -\int\limits^{\sqrt{x}}_0 {\sqrt{t} \sin (t)} \, dt + \int\limits^{x^3}_0 {\sqrt{t} \sin (t)} \, dt \bigg]$
4. Chain Rule [Integration Rule - Fundamental Theorem of Calculus 2]:     $\displaystyle y' = 9 \bigg[ - \bigg( {\sqrt{\sqrt{x}} \sin (\sqrt{x}) \bigg) \cdot \frac{d}{dx}[\sqrt{x}] + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot \frac{d}{dx}[x^3] \bigg]$
5. Basic Power Rule:                                                                                         $\displaystyle y' = 9 \bigg[ - \bigg( {\sqrt{\sqrt{x}} \sin (\sqrt{x}) \bigg) \frac{1}{2\sqrt{x}} + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot 3x^2 \bigg]$
6. Simplify:                                                                                                        $\displaystyle y' = 9 \bigg[ \frac{- \bigg( x^\big{\frac{1}{4}} \sin (\sqrt{x}) \bigg)}{2\sqrt{x}} + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot 3x^2 \bigg]$
7. Rewrite:                                                                                                         $\displaystyle y' = \frac{9 \bigg[ 6x^\big{\frac{9}{4}} \sqrt{x^3} \sin (x^3) - \sin (\sqrt{x}) \bigg] }{2x^\big{\frac{1}{4}}}$

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

Unit: Integration