How would I find dy/dx of y = x^3 ln x ?

Answer 1

Product rule:

If h(x) = f(x) * g(x) then

h'(x) = f'(x)*g(x) + g'(x)*f(x)

Here, y = h(x) = x^3 * ln x
and f(x) = x^3; g(x) = ln x

so dy/dx = h'(x) = 3x^2 * ln x + (1/x) * x^3

=3x^2 * ln x + x^2
=x^2 (3ln x + 1)
=x^2 (ln x^3 + 1)

Answer 2

Use product rule

dy/dx = d(x^3)/dx * ln x + x^3 * d(ln x)/dx

= 3x^2 * ln x + x^3 * 1/x

= 3x^2 ln x + x^2

= x^2 * (3 ln x + 1)

Answer 3

dy/dx=d(x^3 lnx)/dx
=ln(x) d(x^3)/dx + x^3 d(ln[x])/dx
=ln(x) 3x^2 + x^3 1/x
=ln(x) 3x^2 + x^2
=x^2(ln(x)+1)

Answer 4

use product rule and solve as follows::

y=x^3 lnx

dy/dx=d(x^3 ln x)/dx

dy/dx=d(x^3)/dx lnx + d(ln x)/dx

dy/dx=3x^2 ln x + x^2

Answer 5

using product rule

dy/dx = x^3(1/x) + ln(x) * (3x^2)

= x^2 + (3x^2) lnx

Answer 6

y = x^3 ln x
y ' = x^3 * (1/x) + (3x^2) * ln x
y ' = x^2(1 + ln x)
Th

Answer 7

there are many ways to do a differentiation problem. the best way is the chain rule method using substitutions.

Here you may use any,

the simple answer is as follows:
3x^2lnx + x^2

you probably know how

Answer 8

Use the product rule. d(fg(x))/dx = f(x)*dg(x)/dx + df(x)/dx*g(x), or using prime notation, (fg)'(x) = f(x)g'(x)+g(x)f'(x), where f(x) = x^3 and g(x) = ln x.