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# Derivative of 5ex plus 2

Updated: 9/19/2023

Wiki User

12y ago

Best Answer

5ex+2?

d/dx(u+v)=du/dx+dv/dx

d/dx(5ex+2)=d/dx(5ex)+d/dx(2)

-The derivative of 5ex is:

d/dx(cu)=c*du/dx where c is a constant.

d/dx(5ex)=5*d/dx(ex)

-The derivative of 2 is 0 because it is a constant.

d/dx(5ex+2)=(5*d/dx(ex))+(0)

d/dx(5ex+2)=5*d/dx(ex)

-The derivative of ex is:

d/dx(eu)=eu*d/dx(u)

d/dx(ex)=ex*d/dx(x)

d/dx(5ex+2)=5*(ex*d/dx(x))

-The derivative of x is:

d/dx(xn)=nxn-1

d/dx(x)=1*x1-1

d/dx(x)=1*x0

d/dx(x)=1*(1)

d/dx(x)=1

d/dx(5ex+2)=5*(ex*1)

d/dx(5ex+2)=5*(ex)

d/dx(5ex+2)=5ex

5ex+2?

d/dx(cu)=c*du/dx where c is a constant.

d/dx(5ex+2)=5*d/dx(ex+2)

-The derivative of ex+2 is:

d/dx(eu)=eu*d/dx(u)

d/dx(ex+2)=ex+2*d/dx(x+2)

d/dx(5ex+2)=5*(ex+2*d/dx(x+2))

-The derivative of x+2 is:

d/dx(u+v)=du/dx+dv/dx

d/dx(x+2)=d/dx(x)+d/dx(2)

d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]

-The derivative of x is:

d/dx(xn)=nxn-1

d/dx(x)=1*x1-1

d/dx(x)=1*x0

d/dx(x)=1*(1)

d/dx(x)=1

-The derivative of 2 is 0 because it is a constant.

d/dx(5ex+2)=5*[ex+2*(1+0)]

d/dx(5ex+2)=5*[ex+2*(1)]

d/dx(5ex+2)=5*[ex+2]

d/dx(5ex+2)=5ex+2

Wiki User

12y ago
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