   Chapter 9.5, Problem 46E ### 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

# Nerve response The number of action potentials produced by a nerve, t seconds after a stimulus, is given by N ( t ) = 25 t + 4 t 2 + 2 − 2 Find the rate at which the action potentials are produced by the nerve.

To determine

To calculate: The rate at which the action potentials are produces by the nerve if the number of action potentials produced by a nerve, t seconds after a stimulus, is given by the equation N(t)=25t+(4t2+2)2.

Explanation

Given Information:

The number of action potentials produced by a nerve, t seconds after a stimulus, is given by the equation N(t)=25t+(4t2+2)2.

Formula Used:

As per the quotient rule, if two functions are given in the form f(x)g(x), then the derivative is given as:

ddx(fg)=f.gg.fg2.

The rate of the function is the first derivative of the function.

If one function is a sum of multiple functions, then the derivative is:

f(x)=f1(x)+f2(x).

Calculation:

Consider the provided equation, N(t)=25t+(4t2+2)2.

The derivative of the first expression 25t is:

f1(t)=25tf1(t)=25dtdt=25

The derivative of the second expression is:

Consider the provided expression is (4t2+2),

In the expression, the value of,

f(t)=4

And

g(t)=(t2+2)

Apply the quotient rule of the expression,

ddt(fg)=ddt(4).(t2+2)ddt(t2+2).(4)(t2+2)2

Evaluate the expression further with the help of f(x)=f1(x)+f2(x).

ddt(fg)=(ddt(4))

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