   Chapter 4.5, Problem 49E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding an Equation of a Tangent Line In Exercises 45-52, find an equation of the tangent line to the graph of the function at the given point. See Example 5. f ( x ) = ln 5 ( x + 2 ) x ; ( − 5 2 , 0 )

To determine

To calculate: The equation of the tangent to the graph of the function f(x)=ln5(x+2)x at the point (52,0).

Explanation

Given information:

The function is f(x)=ln5(x+2)x and the point is (52,0).

Formula used:

Let u be a differentiable function of x then, ddx[lnx]=1x,x>0 and ddx(lnu)=1ududx,u>0.

Properties of logarithm,

lnab=lnalnbln(ab)=lna+lnb

Where, a and b are any real number or variable.

The derivative of function f(x)=un using the chain rule is,

ddx(un)=nun1dudx

Where, u is the function of x.

Equation of line passes through the point (x1,y1) is given as,

yy1=m(xx1)

Where, m is the slope and m=dydx at (x1,y1).

Calculation:

Consider the function, f(x)=ln5(x+2)x.

Apply the properties of logarithmic to the provided function,

ln5(x+2)x=ln5+ln(x+2)lnx

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