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Calculus Examples
2xcos(x2)2xcos(x2)
Step 1
Since 2 is constant with respect to x, the derivative of 2xcos(x2) with respect to x is 2ddx[xcos(x2)].
2ddx[xcos(x2)]
Step 2
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)=cos(x2).
2(xddx[cos(x2)]+cos(x2)ddx[x])
Step 3
Step 3.1
To apply the Chain Rule, set u as x2.
2(x(ddu[cos(u)]ddx[x2])+cos(x2)ddx[x])
Step 3.2
The derivative of cos(u) with respect to u is -sin(u).
2(x(-sin(u)ddx[x2])+cos(x2)ddx[x])
Step 3.3
Replace all occurrences of u with x2.
2(x(-sin(x2)ddx[x2])+cos(x2)ddx[x])
2(x(-sin(x2)ddx[x2])+cos(x2)ddx[x])
Step 4
Step 4.1
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
2(x(-sin(x2)(2x))+cos(x2)ddx[x])
Step 4.2
Multiply 2 by -1.
2(x(-2sin(x2)x)+cos(x2)ddx[x])
2(x(-2sin(x2)x)+cos(x2)ddx[x])
Step 5
Raise x to the power of 1.
2(x1x(-2sin(x2))+cos(x2)ddx[x])
Step 6
Raise x to the power of 1.
2(x1x1(-2sin(x2))+cos(x2)ddx[x])
Step 7
Use the power rule aman=am+n to combine exponents.
2(x1+1(-2sin(x2))+cos(x2)ddx[x])
Step 8
Add 1 and 1.
2(x2(-2sin(x2))+cos(x2)ddx[x])
Step 9
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
2(x2(-2sin(x2))+cos(x2)⋅1)
Step 10
Multiply cos(x2) by 1.
2(x2(-2sin(x2))+cos(x2))
Step 11
Step 11.1
Apply the distributive property.
2(x2(-2sin(x2)))+2cos(x2)
Step 11.2
Multiply -2 by 2.
-4x2sin(x2)+2cos(x2)
-4x2sin(x2)+2cos(x2)