PLease explain how to differentiate: (3+x)lnx?

Question

Thanks

Answer 1

The product rule states that:
d(uv)/dx = vdu/dx + udv/dx
where u and v are functions of x

Take u = 3+x
v = lnx

d((3+x)lnx)/dx

= lnx*d(3+x)/dx + (3+x)d(lnx)/dx

=lnx[d(3)/dx + d(x)/dx] + (3+x)[d(lnx)/dx]

=lnx(0 + 1) + (3+x)*1/x

= lnx + 3/x + 1

Answer 2

What you have here is a function that is in the form:
f(x) = U(x) * V(x).

Keep in mind that if f(x) = U(x) V(x) then f'(x) = U'(x)V(x) + U(x)V'(x) AND

if f(x) = U(x) + V(x) then f'(x) = U'(x) + V'(x)

To differentiate you have two options:

Option 1: Simplify your function then differentiate:
f(x) = 3lnx + xlnx
f'(x) = 3/x + (1)lnx + x/x = 3/x + lnx + 1 = (3 + xlnx + x) / x

Option 2: You apply the formula for differentiating a product of functions:
f(x) = (3 + x)lnx
f'(x) = (1)lnx + (3+x)/x = (xlnx + 3 + x) / x

Answer 3

y=(3+x)lnx
dy/dx=(3+x)1/x+1lnx by parts.
=1+3/x+lnx.

Answer 4

(3+x)lnx =3lnx + xlnx
dy/dx = 3/x + lnx + 1

The derivative of lnx is 1/x
Use product rule to fid derivative of xlnx
=x*1/x+ 1*lnx = 1+lnx

Answer 5

F(x)=(x+3)LnX
F has 2 sentences .
and will solve in this way :
Diff(First Sentence)*(Second Sentence)+Diff(Second Sentence)*(First Sentence)
F'(x)=(1)(LnX)+(1/x)(X+3)
==>F'(x)=LnX+(X+3)/X
----------------------
Second way (difficulter)
F(x)=XlnX+3LnX
F'(x)=LnX+(1/x)X+3/x
==>F'(x)=LnX+(X+3)/X

Answer 6

let's use the product rule
D(f*g) = Df*g+f*Dg
D[(3+x)*ln x] = D(3+x)*ln x + (3+x)*D(ln x)
=1*ln x + (3+x)*1/x
=ln x +3/x+1

D(3+x)=D3+Dx=0+1
D(ln x)=1/x

Answer 7

the derivative of a function at x is defined as equal to the limit as h approaches zero of [f(x+h)-f(x)]/2h. Every derivative must ultimately solve that equation.