   Chapter 3.4, Problem 54E

Chapter
Section
Textbook Problem

Find an equation of the tangent line to the curve at the given point. y = x e − x 2 ,     ( 0 , 0 )

To determine

To find: The equation of the tangent line to the curve at the point.

Explanation

Given:

The function is y=xex2.

The point is (0,0).

Result used:

Chain Rule

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product

F(x)=g(h(x))h(x) (1)

Product Rule:

If f(x). and g(x) are both differentiable function, then

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (3)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

Calculation:

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y)=ddx(xex2)

Apply the product rule as shown in equation (2)

dydx=xddx[ex2]+ex2ddx[x]=xddx[ex2]+ex2[1x11]=xddx[ex2]+ex2

dydx=xddx[ex2]+ex2 (4)

Obtain the derivative ddx[ex2] by using the chain rule as shown in equation (1).

Let h(x)=x2 and g(u)=eu  where u=h(x).

ddx[ex2]=g(h(x))h(x) (5)

The derivative g(h(x)) is compute as follows,

g(h(x))=g(u)=ddu(g(u))=ddueu=eu

Substitute u=x2 in the above equation,

g(h(x))=ex2

Thus, the derivative g(h(x)) is g(h(x))=ex2

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