Question

Find the derivative of the function

Answer 1

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

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Integration

• Integrals

Integration Rule [Fundamental Theorem of Calculus 2]:
`\displaystyle (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`

Explanation:

Step 1: Define

Identify

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

Step 2: Differentiate

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

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

Unit: Integration