1. Find y' a. y = ln (x^2 + e^{2x} + 1)^3 b. y = sin^{-1} (e^{3x}) c. y = 2 tan^{-1}(4x) 2....


1. Find {eq}y'{/eq}

a. {eq}y = ln (x^2 + e^{2x} + 1)^3 {/eq}

b. {eq}y = sin^{-1} (e^{3x}) {/eq}

c. {eq}y = 2 tan^{-1}(4x) {/eq}

2. Find the point on the graph of {eq}f(x) = x^3 - 12x + 10{/eq} where the tangent line is horizontal.

Slope of the Tangent Plane :

The slope of the tangent line at any given point is {eq}\frac{dy}{dx} {/eq} and for the tangent lines are said to be horizontal, if above derivative equals zero.

Answer and Explanation:


a. {eq}y = ln (x^2 + e^{2x} + 1)^3 {/eq}

On differentiating with respect to x, we get

{eq}y' = \displaystyle \frac{1}{ (x^2 + e^{2x} + 1)^3 }...

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Learn more about this topic:

Tangent Plane to the Surface

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 3

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