Find an equation of the tangent line to the graph

 Calculus Tutor: None Selected Time limit: 1 Day

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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Jul 3rd, 2015

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
May 25th, 2017
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