Answer:
[tex]\displaystyle y' = - \frac{e^{x^2 + 7} \sqrt{\csc 5x} \Bigg[ \bigg[ 5 \cot (5x) - 4x \bigg] \sin (3x + 4) - 6 \cos (3x + 4) \Bigg] }{2}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]:
[tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]:
[tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Derivative Rule [Chain Rule]:
[tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle y = e^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)}[/tex]
Step 2: Differentiate
- Apply Derivative Rule [Product Rule]:
[tex]\displaystyle y' = \big[ e^{x^2 + 7} \big]' \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'[/tex] - Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
[tex]\displaystyle y' = e^{x^2 + 7} (x^2 + 7)' \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'[/tex] - Apply Derivative Rules and Properties [Basic Power Rule + Addition/Subtraction]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'[/tex] - Apply Trigonometric Differentiation [Derivative Rule - Chain Rule]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \cos (3x + 4) (3x + 4)' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'[/tex] - Apply Derivative Rules and Properties [Basic Power Rule + Addition/Subtraction]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'[/tex] - Apply Derivative Rules [Basic Power Rule + Chain Rule]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{\big[ \csc (5x) \big] '}{2\sqrt{\csc (5x)}}[/tex] - Apply Trigonometric Differentiation [Derivative Rule - Chain Rule]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{- \csc (5x) \cot (5x) (5x)'}{2\sqrt{\csc (5x)}}[/tex] - Apply Derivative Rules and Properties [Basic Power Rule + Multiplied Constant]:
[tex]\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{-5 \csc (5x) \cot (5x)}{2\sqrt{\csc (5x)}}[/tex] - Rewrite:
[tex]\displaystyle y' = - \frac{e^{x^2 + 7} \sqrt{\csc 5x} \Bigg[ \bigg[ 5 \cot (5x) - 4x \bigg] \sin (3x + 4) - 6 \cos (3x + 4) \Bigg] }{2}[/tex]
∴ we have found the derivative of the function.
---
Learn more about differentiation: https://brainly.com/question/26836290
Learn more about calculus: https://brainly.com/question/23558817
---
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
