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Calculus Examples
√x3+3x2√x3+3x2
Step 1
Step 1.1
Rewrite x3+3x2 as x2(x+3).
Step 1.1.1
Factor x2 out of x3.
ddx[√x2x+3x2]
Step 1.1.2
Factor x2 out of 3x2.
ddx[√x2x+x2⋅3]
Step 1.1.3
Factor x2 out of x2x+x2⋅3.
ddx[√x2(x+3)]
ddx[√x2(x+3)]
Step 1.2
Pull terms out from under the radical.
ddx[x√x+3]
ddx[x√x+3]
Step 2
Use n√ax=axn to rewrite √x+3 as (x+3)12.
ddx[x(x+3)12]
Step 3
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x and g(x)=(x+3)12.
xddx[(x+3)12]+(x+3)12ddx[x]
Step 4
Step 4.1
To apply the Chain Rule, set u as x+3.
x(ddu[u12]ddx[x+3])+(x+3)12ddx[x]
Step 4.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=12.
x(12u12-1ddx[x+3])+(x+3)12ddx[x]
Step 4.3
Replace all occurrences of u with x+3.
x(12(x+3)12-1ddx[x+3])+(x+3)12ddx[x]
x(12(x+3)12-1ddx[x+3])+(x+3)12ddx[x]
Step 5
To write -1 as a fraction with a common denominator, multiply by 22.
x(12(x+3)12-1⋅22ddx[x+3])+(x+3)12ddx[x]
Step 6
Combine -1 and 22.
x(12(x+3)12+-1⋅22ddx[x+3])+(x+3)12ddx[x]
Step 7
Combine the numerators over the common denominator.
x(12(x+3)1-1⋅22ddx[x+3])+(x+3)12ddx[x]
Step 8
Step 8.1
Multiply -1 by 2.
x(12(x+3)1-22ddx[x+3])+(x+3)12ddx[x]
Step 8.2
Subtract 2 from 1.
x(12(x+3)-12ddx[x+3])+(x+3)12ddx[x]
x(12(x+3)-12ddx[x+3])+(x+3)12ddx[x]
Step 9
Step 9.1
Move the negative in front of the fraction.
x(12(x+3)-12ddx[x+3])+(x+3)12ddx[x]
Step 9.2
Combine 12 and (x+3)-12.
x((x+3)-122ddx[x+3])+(x+3)12ddx[x]
Step 9.3
Move (x+3)-12 to the denominator using the negative exponent rule b-n=1bn.
x(12(x+3)12ddx[x+3])+(x+3)12ddx[x]
Step 9.4
Combine 12(x+3)12 and x.
x2(x+3)12ddx[x+3]+(x+3)12ddx[x]
x2(x+3)12ddx[x+3]+(x+3)12ddx[x]
Step 10
By the Sum Rule, the derivative of x+3 with respect to x is ddx[x]+ddx[3].
x2(x+3)12(ddx[x]+ddx[3])+(x+3)12ddx[x]
Step 11
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
x2(x+3)12(1+ddx[3])+(x+3)12ddx[x]
Step 12
Since 3 is constant with respect to x, the derivative of 3 with respect to x is 0.
x2(x+3)12(1+0)+(x+3)12ddx[x]
Step 13
Step 13.1
Add 1 and 0.
x2(x+3)12⋅1+(x+3)12ddx[x]
Step 13.2
Multiply x2(x+3)12 by 1.
x2(x+3)12+(x+3)12ddx[x]
x2(x+3)12+(x+3)12ddx[x]
Step 14
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
x2(x+3)12+(x+3)12⋅1
Step 15
Multiply (x+3)12 by 1.
x2(x+3)12+(x+3)12
Step 16
To write (x+3)12 as a fraction with a common denominator, multiply by 2(x+3)122(x+3)12.
x2(x+3)12+(x+3)12⋅2(x+3)122(x+3)12
Step 17
Combine (x+3)12 and 2(x+3)122(x+3)12.
x2(x+3)12+(x+3)12(2(x+3)12)2(x+3)12
Step 18
Combine the numerators over the common denominator.
x+(x+3)12(2(x+3)12)2(x+3)12
Step 19
Step 19.1
Move (x+3)12.
x+(x+3)12(x+3)12⋅22(x+3)12
Step 19.2
Use the power rule aman=am+n to combine exponents.
x+(x+3)12+12⋅22(x+3)12
Step 19.3
Combine the numerators over the common denominator.
x+(x+3)1+12⋅22(x+3)12
Step 19.4
Add 1 and 1.
x+(x+3)22⋅22(x+3)12
Step 19.5
Divide 2 by 2.
x+(x+3)1⋅22(x+3)12
x+(x+3)1⋅22(x+3)12
Step 20
Simplify (x+3)1⋅2.
x+(x+3)⋅22(x+3)12
Step 21
Move 2 to the left of x+3.
x+2⋅(x+3)2(x+3)12
Step 22
Step 22.1
Apply the distributive property.
x+2x+2⋅32(x+3)12
Step 22.2
Simplify the numerator.
Step 22.2.1
Multiply 2 by 3.
x+2x+62(x+3)12
Step 22.2.2
Add x and 2x.
3x+62(x+3)12
3x+62(x+3)12
Step 22.3
Factor 3 out of 3x+6.
Step 22.3.1
Factor 3 out of 3x.
3(x)+62(x+3)12
Step 22.3.2
Factor 3 out of 6.
3x+3⋅22(x+3)12
Step 22.3.3
Factor 3 out of 3x+3⋅2.
3(x+2)2(x+3)12
3(x+2)2(x+3)12
3(x+2)2(x+3)12