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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

State each differentiation rule both in symbols and in words.

(a) The Power Rule

(b) The Constant Multiple Rule

(c) The Sum Rule

(d) The Difference Rule

(e) The Product Rule

(f) The Quotient Rule

(g) The Chain Rule

(a)

To determine

To state: The power rule.

Explanation

Statement:

If n is any real number, then ddx(xn)=nxn1.

Example:

Differentiate, f(x)=x2.

Obtain the derivative.

f(x)=ddx(f(x))=ddx(x2)=2x21     (by power rule)=2x

(b)

To determine

To state: The constant multiple rule.

Explanation

Statement:

If c is a constant and f is a differentiable function, then

ddx(cf(x))=cddx(f(x))

Example:

Differentiate, f(x)=3x2.

Obtain the derivative of f(x).

f(x)=ddx(f(x))=ddx(3x2)=3ddx(x2)      (by constant multiple rule)=6x

(c)

To determine

To state: The sum rule.

Explanation

Statement:

If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)+f2(x)]=ddx[f1(x)]+ddx[f2(x)]

Example:

Differentiate, f(x)=x2+2x.

Obtain the derivative of f(x).

ddx[x2+2x]=ddx[x2]+ddx[2x]      (by sum rule)=2x21+2ddx[x]=2x+2[1]=2x+2

(d)

To determine

To state: The difference rule.

Explanation

Statement:

If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=ddx[f1(x)]ddx[f2(x)]

Example:

Differentiate, f(x)=x22x.

Obtain the derivative of f(x).

ddx[x22x]=ddx[x2]ddx[2x]    (by difference rule)=2x212ddx[x]=2x2[1]=2x2

(e)

To determine

To state: The product rule.

Explanation

Statement:

If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

Example:

Differentiate, f(x)=xex.

Obtain the derivative of f(x).

ddx[xex]=xddx[ex]+exddx[x]    (by product rule)=x[ex]+ex[1]=xex+ex

(f)

To determine

To state: The quotient rule.

Explanation

Statement:

If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2(x)]2

Example:

Differentiate of f(x)=exx.

Obtain the derivative of f(x).

ddx[exx]=xddx[ex]exddx[x]x2     (by quotient rule)=x[ex]ex[1]x2=xexexx2

(g)

To determine

To state: The chain rule.

Explanation

Statement:

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

F(t)=f(g(t))g(t)

Example:

Differentiate y=log5(1+2x).

Obtain the derivative of y.

y=ddx(y)=ddx(log5(1+2x))

Let h(x)=1+2x and f(u)=log5u  where u=h(x).

Apply the chain rule as shown in equation (1),

y=f(h(x))h(x) (2)

The derivative f(h(x)) is computed as follows,

f(h(x))=f(u)            [Qu=h(x)]=ddu(f(u))=ddu(log5u)=1uln5             (Qddx(logax)=1xlna)

Substitute u=1+2x in the above equation,

f(h(x))=1(1+2x)ln5

Thus, the derivative f(h(x)) is f(h(x))=1(1+2x)ln5.

The derivative of h(x) is computed as follows,

h(x)=ddx(1+2x)=(0+2(1)x11)=2

Thus, the derivative of h(x) is h(x)=2.

Substitute 1(1+2x)ln5 for f(h(x)) and 2 for h(x) in equation (2),

y=1(1+2x)ln5(2)=2(1+2x)ln5

Therefore, the derivative of y=log5(1+2x) is y=2(1+2x)ln5_.

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P-69RECh-3 P-70RECh-3 P-71RECh-3 P-72RECh-3 P-73RECh-3 P-74RECh-3 P-75RECh-3 P-76RECh-3 P-77RECh-3 P-78RECh-3 P-79RECh-3 P-80RECh-3 P-81RECh-3 P-82RECh-3 P-83RECh-3 P-84RECh-3 P-85RECh-3 P-86RECh-3 P-87RECh-3 P-88RECh-3 P-89RECh-3 P-90RECh-3 P-91RECh-3 P-92RECh-3 P-93RECh-3 P-94RECh-3 P-95RECh-3 P-96RECh-3 P-97RECh-3 P-98RECh-3 P-99RECh-3 P-100RECh-3 P-101RECh-3 P-102RECh-3 P-103RECh-3 P-104RECh-3 P-105RECh-3 P-106RECh-3 P-107RECh-3 P-108RECh-3 P-109RECh-3 P-110RECh-3 P-111RECh-3 P-112RECh-3 P-1PCh-3 P-2PCh-3 P-3PCh-3 P-4PCh-3 P-5PCh-3 P-6PCh-3 P-7PCh-3 P-8PCh-3 P-9PCh-3 P-10PCh-3 P-11PCh-3 P-12PCh-3 P-13PCh-3 P-14PCh-3 P-15PCh-3 P-16PCh-3 P-17PCh-3 P-18PCh-3 P-19PCh-3 P-20PCh-3 P-21PCh-3 P-22PCh-3 P-23PCh-3 P-24PCh-3 P-25PCh-3 P-27PCh-3 P-28PCh-3 P-29PCh-3 P-30PCh-3 P-31PCh-3 P-32PCh-3 P-33PCh-3 P-34PCh-3 P-35P

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