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Chapter 2: Q. 88 (page 199)

Use the definition of the derivative to prove the following special case of the product rule

ddx(x2f(x))=2xf(x)+x2f'(x)

Short Answer

Expert verified

We proved the special case of product function using the definition of the derivative

Step by step solution

01

Given information

We are given a functionddx(x2f(x))=2xf(x)+x2f'(x)

02

Find the derivative

Consider g(x)=x2f(x)

Using the definition of derivative we get,

limh→0g(x+h)-g(x)hlimh→0(x+h)2f(x+h)-x2f(x)hlimh→0(x2+2xh+h2)f(x+h)-x2f(x)hlimh→0x2(f(x+h)-f(x))h+limh→0h(2x+h)f(x+h)h=x2f'(x)+2xf(x)

Hence proved

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Most popular questions from this chapter

Suppose f is a polynomial of degree n and let k be some integer with 0≤k≤n. Prove that if f(x) is of the form f(x)=anxn+an-1xn-1+....+a1x+a0

Then ak=fk(0)k!where fk(x)is the k-th derivative off(x)andk!=k·(k-1)·(k-2)....2·1

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.

f(x)=53x4-13+3x-1100

The following reciprocal rules tells us hoe to differentiate the reciprocal of a function

ddx(1f(x))=-1[f(x)]2

Prove this using

a) definition of the derivative

b) by using the quotient rule

For each function f that follows find all the x-values in the domain of f for which f'(x)=0and all the values for which f'(x)does not exist in later section we will call these values the critical points of f

localid="1648604345877" a)f(x)=x3-2xb)f(x)=x-xc) f(x)=11+xd)f(x)=x2(x-1)(x-2)2

Use (a) the h→0definition of the derivative and then

(b) the z→cdefinition of the derivative to find f'(c) for each function f and value x=c in Exercises 23–38.

26.f(x)=1x2,x=2
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