How do you find the derivative of #5(x^2 + 5)^4(2x)(x − 3)4 + (x^2 + 5)^5(4)(x − 3)^3#?

1 Answer
Jul 16, 2015

Answer:

Like Saikiran Reddy, I will assume that there is an error in the question and we want the derivative of: #5(x^2+5)^4(2x)(x-3)^4 + (x^2+5)^5(4)(x-3)^3 #

Explanation:

I make the assumption because with the correction this is the derivative of #(x^2+5)^5(x-3)^4#. A nice problem for the product and chain rules.

I will start by rewriting the expression:

#5(x^2+5)^4(2x)(x-3)^4 + (x^2+5)^5(4)(x-3)^3#

# = 10x(x^2+5)^4(x-3)^4+4(x^2+5)^5(x-3)^3#

This is a sum of two terms. Let's take out common factors.

#= 2(x^2+5)^4(x-3)^3[5x(x-3) + 2(x^2+5)]#

# = 2(x^2+5)^4(x-3)^3[5x^2-15x + 2x^2+10]#

# = 2(x^2+5)^4(x-3)^3(7x^2-15x+10)#

Now we can differentiate using the product rule for three factors. (The constant #2# will just hang out in front.)

#d/dx(fgh) = f'gh+fg'h+fgh'#

(You can get this formula using the product rule twice. And it's easy enough to remember: the prime just makes its way through the factors one by one.)

In this problem we'll have:
#d/dx(2fgh) = 2f'gh+2fg'h+2fgh'#

So the derivative of our expression is:

#2[4(x^2+5)^3(2x) ] (x-3)^3(7x^2-15x+10)#

# + 2(x^2+5)^4 [ 3(x-3)^2] (7x^2-15x+10)#

# + 2(x^2+5)^4(x-3)^3 [14x-15]#

We can simplify by first simplifying each term:

#16x(x^2+5)^3(x-3)^3(7x^2-15x+10)#

# + 6(x^2+5)^4 (x-3)^2 (7x^2-15x+10)#

# + 2(x^2+5)^4(x-3)^3 (14x-15)#

And now we can remove common factors as we did before differentiating:

#2(x^2+5)^3(x-3)^2[8x(x-3)(7x^2-15x+10) + 3(x^2+5)(7x^2-15x+10)+(x^2+5)(x-3)(14x-15)] ]#

The expression in brackets simplifies (by WolframAlpha) to
#91x^4-390x^3+690x^2-750x+375#

So we end up with:

#2(x^2+5)^3(x-3)^2(91x^4-390x^3+690x^2-750x+375)#

Notes

1
The derivative of #fgh# is:

#d/dx(fgh) = d/dx((fg)h) = (fg)'h+(fg)h'#

# = (f'g+fg')h+fgh'#

# = f'gh+fg'h+fgh'#

2
I've been doing mathematics since the 1970s. I don't need to practice my algebra. By having Wolfram simplify, I can answer more questions on how to do things. :-)