##### Find an equation of the tangent line to the graph

label Calculus
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Find an equation of the tangent line to the graph of y = ln(x2) at the point (5, ln(25)).

Jul 3rd, 2015

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$y=ln(x^2)$

Differentiate y w.r.t. x

$\\ \frac{dy}{dx}=\frac{d}{dx}\big(ln(x^2)\big)\\ \\ Differentiate\hspace{5}using\hspace{5}chain\hspace{5}rule\\ \\ \frac{dy}{dx}=\frac{1}{x^2}\times\frac{d}{dx}\big(x^2\big)\\ \\ \frac{dy}{dx}=\frac{1}{x^2}\times2x\\ \\ \frac{dy}{dx}=\frac{2}{x}\\$

2/x is the slope of the tangent line at any point on the curve.

At (5, ln(25)), slope is

$\\ \frac{dy}{dx}=\frac{2}{5}=0.4\\$

We can use the point slope form of a line to find the equation of the tangent.

Point slope form of a line is

$y-y_0=m(x-x_0)$

where m is the slope of the line

and $(x_0,y_0)$ is a point on the line.

Here, m = 0.4 and $(x_0,y_0)$ = (5, ln(25))

So, equation of tangent line is

$\\ y-ln(25)=0.4(x-5)\\ \\ y-3.22=0.4x-2\\ \\ y=0.4x-2+3.22\\ \\ y=0.4x+1.22$

ANSWER:  y = 0.4x + 1.22

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Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 3rd, 2015

Hi this answer is incorrect

Jul 3rd, 2015

could you have made a wrong calculation

Jul 3rd, 2015

Do you submit the answer online? If yes does the question mention any particular way of giving the answer.

Jul 3rd, 2015

I just checked by plotting the graph and it seems to be correct.

Jul 3rd, 2015
graph.png

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Jul 3rd, 2015
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Jul 3rd, 2015
Sep 24th, 2017
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