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Calculus Examples
f(x)=2x3f(x)=2x3
Step 1
Consider the limit definition of the derivative.
f′(x)=limh→0f(x+h)-f(x)h
Step 2
Step 2.1
Evaluate the function at x=x+h.
Step 2.1.1
Replace the variable x with x+h in the expression.
f(x+h)=2(x+h)3
Step 2.1.2
Simplify the result.
Step 2.1.2.1
Use the Binomial Theorem.
f(x+h)=2(x3+3x2h+3xh2+h3)
Step 2.1.2.2
Apply the distributive property.
f(x+h)=2x3+2(3x2h)+2(3xh2)+2h3
Step 2.1.2.3
Simplify.
Step 2.1.2.3.1
Multiply 3 by 2.
f(x+h)=2x3+6(x2h)+2(3xh2)+2h3
Step 2.1.2.3.2
Multiply 3 by 2.
f(x+h)=2x3+6(x2h)+6(xh2)+2h3
f(x+h)=2x3+6(x2h)+6(xh2)+2h3
Step 2.1.2.4
Remove parentheses.
f(x+h)=2x3+6x2h+6xh2+2h3
Step 2.1.2.5
The final answer is 2x3+6x2h+6xh2+2h3.
2x3+6x2h+6xh2+2h3
2x3+6x2h+6xh2+2h3
2x3+6x2h+6xh2+2h3
Step 2.2
Reorder.
Step 2.2.1
Move x2.
2x3+6hx2+6xh2+2h3
Step 2.2.2
Move x.
2x3+6hx2+6h2x+2h3
Step 2.2.3
Move 2x3.
6hx2+6h2x+2h3+2x3
Step 2.2.4
Move 6hx2.
6h2x+2h3+6hx2+2x3
Step 2.2.5
Reorder 6h2x and 2h3.
2h3+6h2x+6hx2+2x3
2h3+6h2x+6hx2+2x3
Step 2.3
Find the components of the definition.
f(x+h)=2h3+6h2x+6hx2+2x3
f(x)=2x3
f(x+h)=2h3+6h2x+6hx2+2x3
f(x)=2x3
Step 3
Plug in the components.
f′(x)=limh→02h3+6h2x+6hx2+2x3-(2x3)h
Step 4
Step 4.1
Simplify the numerator.
Step 4.1.1
Multiply -1 by 2.
f′(x)=limh→02h3+6h2x+6hx2+2x3-2x3h
Step 4.1.2
Subtract 2x3 from 2x3.
f′(x)=limh→02h3+6h2x+6hx2+0h
Step 4.1.3
Add 2h3+6h2x+6hx2 and 0.
f′(x)=limh→02h3+6h2x+6hx2h
Step 4.1.4
Factor 2h out of 2h3+6h2x+6hx2.
Step 4.1.4.1
Factor 2h out of 2h3.
f′(x)=limh→02h⋅h2+6h2x+6hx2h
Step 4.1.4.2
Factor 2h out of 6h2x.
f′(x)=limh→02h⋅h2+2h(3hx)+6hx2h
Step 4.1.4.3
Factor 2h out of 6hx2.
f′(x)=limh→02h⋅h2+2h(3hx)+2h(3x2)h
Step 4.1.4.4
Factor 2h out of 2h⋅h2+2h(3hx).
f′(x)=limh→02h(h2+3hx)+2h(3x2)h
Step 4.1.4.5
Factor 2h out of 2h(h2+3hx)+2h(3x2).
f′(x)=limh→02h(h2+3hx+3x2)h
f′(x)=limh→02h(h2+3hx+3x2)h
f′(x)=limh→02h(h2+3hx+3x2)h
Step 4.2
Simplify terms.
Step 4.2.1
Cancel the common factor of h.
Step 4.2.1.1
Cancel the common factor.
f′(x)=limh→02h(h2+3hx+3x2)h
Step 4.2.1.2
Divide 2(h2+3hx+3x2) by 1.
f′(x)=limh→02(h2+3hx+3x2)
f′(x)=limh→02(h2+3hx+3x2)
Step 4.2.2
Apply the distributive property.
f′(x)=limh→02h2+2(3hx)+2(3x2)
f′(x)=limh→02h2+2(3hx)+2(3x2)
Step 4.3
Simplify.
Step 4.3.1
Multiply 3 by 2.
f′(x)=limh→02h2+6(hx)+2(3x2)
Step 4.3.2
Multiply 3 by 2.
f′(x)=limh→02h2+6(hx)+6x2
f′(x)=limh→02h2+6(hx)+6x2
Step 4.4
Move h.
f′(x)=limh→02h2+6xh+6x2
Step 4.5
Move 2h2.
f′(x)=limh→06xh+6x2+2h2
Step 4.6
Reorder 6xh and 6x2.
f′(x)=limh→06x2+6xh+2h2
f′(x)=limh→06x2+6xh+2h2
Step 5
Split the limit using the Sum of Limits Rule on the limit as h approaches 0.
limh→06x2+limh→06xh+limh→02h2
Step 6
Evaluate the limit of 6x2 which is constant as h approaches 0.
6x2+limh→06xh+limh→02h2
Step 7
Move the term 6x outside of the limit because it is constant with respect to h.
6x2+6xlimh→0h+limh→02h2
Step 8
Move the term 2 outside of the limit because it is constant with respect to h.
6x2+6xlimh→0h+2limh→0h2
Step 9
Move the exponent 2 from h2 outside the limit using the Limits Power Rule.
6x2+6xlimh→0h+2(limh→0h)2
Step 10
Step 10.1
Evaluate the limit of h by plugging in 0 for h.
6x2+6x⋅0+2(limh→0h)2
Step 10.2
Evaluate the limit of h by plugging in 0 for h.
6x2+6x⋅0+2⋅02
6x2+6x⋅0+2⋅02
Step 11
Step 11.1
Simplify each term.
Step 11.1.1
Multiply 6x⋅0.
Step 11.1.1.1
Multiply 0 by 6.
6x2+0x+2⋅02
Step 11.1.1.2
Multiply 0 by x.
6x2+0+2⋅02
6x2+0+2⋅02
Step 11.1.2
Raising 0 to any positive power yields 0.
6x2+0+2⋅0
Step 11.1.3
Multiply 2 by 0.
6x2+0+0
6x2+0+0
Step 11.2
Combine the opposite terms in 6x2+0+0.
Step 11.2.1
Add 6x2 and 0.
6x2+0
Step 11.2.2
Add 6x2 and 0.
6x2
6x2
6x2
Step 12