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Find an equation of the tangent line to the curve y = sqrt x at (36, 6).

Question:

Find an equation of the tangent line to the curve {eq}y = \sqrt x {/eq} at (36, 6).

Equation of Tangent Line:

Suppose {eq}f(x) {/eq} is a differentiable function. Then {eq}f'(a) {/eq} is the slope of the line tangent to {eq}f(x) {/eq} at {eq}(a,f(a)){/eq}.

Combining with the point-slope form of a line we have the equation of the line tangent to {eq}f(x) {/eq} at {eq}x=a {/eq} as

{eq}y-f(a)=f'(a)(x-a). {/eq}

Answer and Explanation:

The derivative of {eq}y=\sqrt{x} {/eq} is

{eq}\begin{align} y'&=\dfrac{d}{dx}[\sqrt{x} ]\\ &=\dfrac{1}{2\sqrt{x}}. \end{align} {/eq}

So the slope of the tangent line at {eq}(36,6) {/eq} is

{eq}m=y'(36)=\dfrac{1}{2\sqrt{36}}=\dfrac{1}{12} {/eq}

So the equation of the tangent line is

{eq}y-6=\frac{1}{12}(x-6). {/eq}


Learn more about this topic:

Slopes of Tangent & Secant Lines

from TExES Mathematics 4-8 Exam (115): Study Guide & Review

Chapter 18 / Lesson 6
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