How do you differentiate #y= e^(2x) (x^2 - 4) ln x#?

1 Answer
Jul 30, 2015

Answer:

#y = e^(2x)/x * [ 2lnx * (x^3 + x^2 - 4x) + x^2 - 4]#

Explanation:

Start by rewriting your function as

#y = e^(2x) * x^2 * lnx - 4 * e^(2x) * lnx#

You can use the sum rule to write

#d/dx(y) = underbrace(d/dx(e^(2x) * x^2 * lnx))_(color(red)(d/dx(f))) - 4 * underbrace(d/dx(e^(2x) * lnx))_(color(red)(d/dx(g)))#

Let's start with the second one because it is easier to calculate. For functions that can be written as the product of two other functions,

#y = f(x) * g(x)#,

the product rule will allow you to calculate their derivative by

#color(blue)(d/dx(y) = f^'(x) * g(x) + f(x) * g^'(x)#

In your case, you can calculate #d/dx(e^(2x) * lnx) = d/dx(g)# by

#g^' = d/dx(e^(2x)) * lnx + e^(2x) * d/dx(lnx)#

You can calculate #d/dx(e^(2x))# by using the chain rule, for

#z(u) = e^u# and #u = 2x#, so that

#d/dx(z) = d/(du)(z) * d/dx(u)#

#d/dx(z) = d/(du)e^u * d/dx(2x)#

#d/dx(z) = e^u * 2 = 2 * e^(2x)#

This means that you have

#g^' = 2 * e^(2x) * lnx + e^(2x) * 1/x#

#g^' = 3^(2x)(2lnx + 1/x)#

Now move to #d/dx(f)#, which can be written as the product of 3 functions.

The general form of the product rule for such functions is

#d/dx(a * b * c) = d/dx(a*b) * c + (a * b) * d/dx(c)#

#d/dx(a * b * c) = [d/dx(a) * b + a * d/dx(b)] * c + (a * b) * d/dx(c)#

#color(blue)(d/dx(a * b * c) = [d/dx(a)] * bc + a [d/dx(b)]c + ab[d/dx(c)])#

So, in your case, you have

#f^' = [d/dx(e^(2x))] * x^2 * lnx + e^(2x) * [d/dx(x^2)] * lnx + e^(2x) * x^2 * d/dx(lnx)#

#f^' = 2 * e^(2x) * x^2 * lnx + 2x * e^(2x) * lnx + e^(2x) * x^(color(red)(cancel(color(black)(2)))) * 1/color(red)(cancel(color(black)(x)))#

#f^' = x * e^(2x) (2xlnx + 2lnx + 1)#

Your target derivative will thus be

#y^' = f^' - 4g^'#

#y^' = 2 * e^(2x) * x^2 * lnx + 2x * e^(2x) * lnx + e^(2x) * x - 4 * (2 * e^(2x) * lnx + e^(2x)/x)#

#y^' = 2 * e^(2x) * x^2 * lnx + 2x * e^(2x) * lnx + e^(2x) * x - 8e^(2x) * lnx - 4/xe^(2x)#

#y^' = e^(2x)/x * (2x^3 * lnx + 2x^2 * lnx + x^2 - 8x * lnx - 4)#

Finally, you get

#y^' = color(green)(e^(2x)/x * [ 2lnx * (x^3 + x^2 - 4x) + x^2 - 4])#

SIDE NOTE You could just as easily have used the product rule for three functions for your starting function, which would have resulted in

#y^' = [d/dx(e^(2x))] * (x^2 - 4) * lnx + e^(2x) * [d/dx(x^2 - 4)] * lnx + e^(2x) * (x^2 - 4) * d/dx(lnx)#