How do you find the derivative of 5(x^2 + 5)^4(2x)(x − 3)4 + (x^2 + 5)^5(4)(x − 3)^35(x2+5)4(2x)(x−3)4+(x2+5)5(4)(x−3)3?
1 Answer
Like Saikiran Reddy, I will assume that there is an error in the question and we want the derivative of:
Explanation:
I make the assumption because with the correction this is the derivative of
I will start by rewriting the expression:
= 10x(x^2+5)^4(x-3)^4+4(x^2+5)^5(x-3)^3=10x(x2+5)4(x−3)4+4(x2+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(x2+5)4(x−3)3[5x(x−3)+2(x2+5)]
= 2(x^2+5)^4(x-3)^3[5x^2-15x + 2x^2+10]=2(x2+5)4(x−3)3[5x2−15x+2x2+10]
= 2(x^2+5)^4(x-3)^3(7x^2-15x+10)=2(x2+5)4(x−3)3(7x2−15x+10)
Now we can differentiate using the product rule for three factors. (The constant
(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:
So the derivative of our expression is:
+ 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:
+ 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:
The expression in brackets simplifies (by WolframAlpha) to
So we end up with:
Notes
1
The derivative of
= (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. :-)