Question

Differentiate the following with respect to xirrelevant answers will be reported​

Answer 1

Answer:

$\displaystyle y' = - \frac{e^{x^2 + 7} \sqrt{\csc 5x} \Bigg[ \bigg[ 5 \cot (5x) - 4x \bigg] \sin (3x + 4) - 6 \cos (3x + 4) \Bigg] }{2}$

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 [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

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

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle y = e^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)}$

Step 2: Differentiate

1. Apply Derivative Rule [Product Rule]:
   $\displaystyle y' = \big[ e^{x^2 + 7} \big]' \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'$
2. Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
   $\displaystyle y' = e^{x^2 + 7} (x^2 + 7)' \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'$
3. Apply Derivative Rules and Properties [Basic Power Rule + Addition/Subtraction]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \big[ \sin (3x + 4) \big]' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'$
4. Apply Trigonometric Differentiation [Derivative Rule - Chain Rule]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \cos (3x + 4) (3x + 4)' \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'$
5. Apply Derivative Rules and Properties [Basic Power Rule + Addition/Subtraction]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \big[ \sqrt{\csc (5x)} \big]'$
6. Apply Derivative Rules [Basic Power Rule + Chain Rule]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{\big[ \csc (5x) \big] '}{2\sqrt{\csc (5x)}}$
7. Apply Trigonometric Differentiation [Derivative Rule - Chain Rule]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{- \csc (5x) \cot (5x) (5x)'}{2\sqrt{\csc (5x)}}$
8. Apply Derivative Rules and Properties [Basic Power Rule + Multiplied Constant]:
   $\displaystyle y' = 2xe^{x^2 + 7} \sin (3x + 4) \sqrt{\csc (5x)} + 3e^{x^2 + 7} \cos (3x + 4) \sqrt{\csc (5x)} + e^{x^2 + 7} \sin (3x + 4) \frac{-5 \csc (5x) \cot (5x)}{2\sqrt{\csc (5x)}}$
9. Rewrite:
   $\displaystyle y' = - \frac{e^{x^2 + 7} \sqrt{\csc 5x} \Bigg[ \bigg[ 5 \cot (5x) - 4x \bigg] \sin (3x + 4) - 6 \cos (3x + 4) \Bigg] }{2}$

∴ we have found the derivative of the function.

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Learn more about differentiation: brainly.com/question/26836290
Learn more about calculus: brainly.com/question/23558817

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

Unit: Differentiation