determine $\frac{dy}{dx}$ for $y = \cos^{-1}(9x^4)$.

answer

$\frac{dy}{dx} =$

Question

determine $\frac{dy}{dx}$ for $y = \cos^{-1}(9x^4)$.

answer

$\frac{dy}{dx} =$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Inverse Trigonometric Derivatives", "Chain Rule Differentiation", "Power Rule Differentiation" ], "new_concepts": [], "current_concepts": [ "Inverse Trigonometric Derivatives", "Chain Rule Differentiation", "Power Rule Differentiation" ] } ## Identify the outer and inner functions $$ \begin{aligned} y &= f(u) = \cos^{-1}(u) \ u &= g(x) = 9x^4 \end{aligned} $$ ## Differentiate the outer and inner functions $$ \begin{aligned} \frac{dy}{du} &= -\frac{1}{\sqrt{1 - u^2}} \ \frac{du}{dx} &= 36x^3 \end{aligned} $$ ## Apply the chain rule to find the derivative $$ \begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \ \frac{dy}{dx} &= -\frac{1}{\sqrt{1 - (9x^4)^2}} \cdot 36x^3 \ \frac{dy}{dx} &= -\frac{36x^3}{\sqrt{1 - 81x^8}} \end{aligned} $$ Determine $\frac{dy}{dx}$ for $y = \cos^{-1}(9x^4)$. $\frac{dy}{dx} =$ $-\frac{36x^3}{\sqrt{1 - 81x^8}}$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Inverse Trigonometric Derivatives" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the outer and inner functions

Differentiate the outer and inner functions

Apply the chain rule to find the derivative

Answer: