`h(u) = Au^3 + Bu^2 + Cu` Differentiate the function

Textbook Question

Chapter 3, 3.1 - Problem 21 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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a-maths-9's profile pic

a-maths-9 | Middle School Teacher | (Level 1) Adjunct Educator

Posted on

we will differentiate w.r.t u

we will use power rule

`f'(x)=d/(du)(Au^3+Bu^2+Cu)`

`f'(x)=3Au^2+2Bu+C` 

kommalapativamsi46's profile pic

kommalapativamsi46 | Elementary School Teacher | (Level 1) Adjunct Educator

Posted on

Given h(u) = A*(u^3) + B*(u^2) + C*u

Differentiate with respect to u on both sides

h'(u) = 3 A*(u^2) + 2 B*(u) + C

       Thus we can get the first derivative of h(u)

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hkj1385 | (Level 1) Assistant Educator

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NOTE: 1) If f(u) = u^n  ; where n = integer; then 

f'(u) = d[f(u)]/du = n*u^(n-1)  

2) If f(u) = k ; where 'k' = real number, then

f'(u) = d[f(u)]/du = 0

Now, given equation of curve:-

h(u) = A*(u^3) + B*(u^2) + C*u

Differentiating h(u) with respect to 'u' we get

h'(u) = 3A*(u^2) + 2B*(u^1) + C*(u^0)

or, h'(u) = 3A*(u^2) + 2B*u + C

joker5's profile pic

joker5 | (Level 1) eNoter

Posted on

To differentiate the function, you need to use the Power Rule, which states:

if `y=kx^(m) `

then `y'=(km)x^(m+1) `

So:

if h(u) = A(u^3) + B(u^2) + Cu

then h'(u) = 3A(u^3-1) + 2B(u^2-1) + C(u^1-0)

Simplifying, your final answer is h'(u) = 3A(u^2) + 2Bu + C

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kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

`h(u) = Au^3 + Bu^2 + Cu`

so,

`h'(u) = A* 3*u^2 + 2*Bu + C`

       ` = 3Au^2 +2Bu +C`

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