   Chapter 9.6, Problem 28E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 25-28, write the equation of the line tangent to the graph of each function at the indicated point. As a check, graph both the function and the tangent line you found to see whether it looks correct. 26 .  y  =( x 2 +1) 3 at  ( 2, 125 )

To determine

To calculate: The equation of line tangent to the function y=(x2+1)3 at the point (2,125) and check by plotting the graph of the tangent line and the function.

Explanation

Given Information:

The function is y=(x2+1)3 and the line is tangent at the point (2,125).

Formula used:

Power rule:

If y=un and u is differentiable function of x, then,

dydx=nun1dudx

According to the property of differentiation, if a function is of the form f(x)=u(x)+v(x), then,

f(x)=u(x)+v(x)

According to the product rule, if f(x)=u(x)v(x), then,

f(x)=u(x)v(x)+v(x)u(x)

According to the point slope form, the equation of a line with slope m and point (x1,y1) is,

yy1=m(xx1)

Calculation:

Consider the provided function,

y=(x2+1)3

The equation of line tangent to the equation is found using the point-slope form.

The slope of tangent is found by differentiating the function.

Now, consider (x2+1) to be u,

y=u3

Differentiate both sides with respect to x,

dydx=ddx(u3)

Simplify using the power rule,

dydx=3u31dudx=3u2dudx

Substitute (x2+1) for u,

dydx=3(x2+1)2ddx(x2+1)=3(x2+1)2(ddx<

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