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Calculus Examples
13(x2+2)3213(x2+2)32
Step 1
Since 1313 is constant with respect to xx, the derivative of 13(x2+2)3213(x2+2)32 with respect to xx is 13ddx[(x2+2)32]13ddx[(x2+2)32].
13ddx[(x2+2)32]13ddx[(x2+2)32]
Step 2
Step 2.1
To apply the Chain Rule, set u as x2+2.
13(ddu[u32]ddx[x2+2])
Step 2.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=32.
13(32u32-1ddx[x2+2])
Step 2.3
Replace all occurrences of u with x2+2.
13(32(x2+2)32-1ddx[x2+2])
13(32(x2+2)32-1ddx[x2+2])
Step 3
To write -1 as a fraction with a common denominator, multiply by 22.
13(32(x2+2)32-1⋅22ddx[x2+2])
Step 4
Combine -1 and 22.
13(32(x2+2)32+-1⋅22ddx[x2+2])
Step 5
Combine the numerators over the common denominator.
13(32(x2+2)3-1⋅22ddx[x2+2])
Step 6
Step 6.1
Multiply -1 by 2.
13(32(x2+2)3-22ddx[x2+2])
Step 6.2
Subtract 2 from 3.
13(32(x2+2)12ddx[x2+2])
13(32(x2+2)12ddx[x2+2])
Step 7
Combine 32 and (x2+2)12.
13(3(x2+2)122ddx[x2+2])
Step 8
Multiply 3(x2+2)122 by 13.
3(x2+2)122⋅3ddx[x2+2]
Step 9
Multiply 2 by 3.
3(x2+2)126ddx[x2+2]
Step 10
Factor 3 out of 3(x2+2)12.
3((x2+2)12)6ddx[x2+2]
Step 11
Step 11.1
Factor 3 out of 6.
3(x2+2)123⋅2ddx[x2+2]
Step 11.2
Cancel the common factor.
3(x2+2)123⋅2ddx[x2+2]
Step 11.3
Rewrite the expression.
(x2+2)122ddx[x2+2]
(x2+2)122ddx[x2+2]
Step 12
By the Sum Rule, the derivative of x2+2 with respect to x is ddx[x2]+ddx[2].
(x2+2)122(ddx[x2]+ddx[2])
Step 13
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
(x2+2)122(2x+ddx[2])
Step 14
Since 2 is constant with respect to x, the derivative of 2 with respect to x is 0.
(x2+2)122(2x+0)
Step 15
Step 15.1
Add 2x and 0.
(x2+2)122(2x)
Step 15.2
Combine 2 and (x2+2)122.
2(x2+2)122x
Step 15.3
Combine 2(x2+2)122 and x.
2(x2+2)12x2
Step 15.4
Cancel the common factor.
2(x2+2)12x2
Step 15.5
Simplify the expression.
Step 15.5.1
Divide (x2+2)12x by 1.
(x2+2)12x
Step 15.5.2
Reorder the factors of (x2+2)12x.
x(x2+2)12
x(x2+2)12
x(x2+2)12