   Chapter 11.3, Problem 55E Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

In Exercises 29-56, calculate d y d x . You need not expand your answers. [HINT: See Example 1 and 2.] y = ( x + 3 ) ( x + 1 ) ( x + 2 ) 3 x − 1

To determine

To calculate: The derivative of function y=(x+3)(x+1)(x+2)3x1.

Explanation

Given Information:

The provided function is y=(x+3)(x+1)(x+2)3x1.

Formula used:

Quotient rule of derivative of differentiable functions, f(x) and g(x) is,

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2 where, g(x)0.

Product rule of derivative of differentiable functions, f, g and h is,

(fgh)=fgh+fgh+fgh

The derivative of a constant is 0.

Derivative of function using power rule, y=xn is dydx=nxn1.

Constant multiple rule of derivative of function f(x) is ddx[cf(x)]=cddx[f(x)] where, c is constant.

Sum and Difference rule of derivative is,

ddx[f(x)±g(x)]=ddx[f(x)]±ddx[g(x)] where, f(x) and g(x) are any two differentiable functions.

Calculation:

Consider the provided function,

y=(x+3)(x+1)(x+2)3x1

To find the derivative of the function, apply the quotient rule,

dydx=ddx[(x+3)(x+1)(x+2)](3x1)(x+3)(x+1)(x+2)ddx[3x1](3x1)2

Apply product rule to the product (x+3)(x+1)(x+2),

ddx[(x+3)(x+1)(x+2)]=[ddx(x+3){(x+2)(x+1)}+ddx(x+1){(x+2)(x+3)}+ddx(x+2){(x+1)(x+3)}]

Apply power rule and constant derivative rule,

ddx[(x+3)(x+1)(x+2)]=[{1x11+0}(x+1)(x+2)+{1x11+0}(x+2)(x+3)+{1x11+0}(x+3)(x+1)]=[(x0)(x+1)(x+2)+(x0)(x+3)(x+2)+(x0)(x+3)(x+1)]=[1(x+1)(x+2)+1(x+3)(x+2)+1(x+3)(x+1)]=[(x2+x+2x+2)+(x2+3x+2x+6)+(x2+3x+x+3)]

On further simplification,

ddx[

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