Question

Find the derivative of the function $f(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})$. $f'(x) = 12 \sec^2(\sqrt[3]{x^2 - 4x})(x^2 - 4x)^{-\frac{2}{3}}(2x - 4)$ $f'(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})$ $f'(x) = \sec^2(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})(2x - 4)$ $f'(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})(x^2 - 4x)^{-\frac{2}{3}}(2x - 4)$

          Find the derivative of the function $f(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})$.
$f'(x) = 12 \sec^2(\sqrt[3]{x^2 - 4x})(x^2 - 4x)^{-\frac{2}{3}}(2x - 4)$
$f'(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})$
$f'(x) = \sec^2(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})(2x - 4)$
$f'(x) = 4 \sec^3(\sqrt[3]{x^2 - 4x})\tan(\sqrt[3]{x^2 - 4x})(x^2 - 4x)^{-\frac{2}{3}}(2x - 4)$

Find the derivative of the function f(x) = 4 sec^3(√(x^2 - 4x)).
f'(x) = 12 sec^2(√(x^2 - 4x))(x^2 - 4x)^-(2)/(3)(2x - 4)
f'(x) = 4 sec^3(√(x^2 - 4x))tan(√(x^2 - 4x))
f'(x) = sec^2(√(x^2 - 4x))tan(√(x^2 - 4x))(2x - 4)
f'(x) = 4 sec^3(√(x^2 - 4x))tan(√(x^2 - 4x))(x^2 - 4x)^-(2)/(3)(2x - 4)

Added by Juan R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Avinash Vishwakarma

06/02/2022

Step 1

Step 1: Given function f(x) = 4sec^3(∛(x^2) - 4x) Show more…

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