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

Jul 16, 2015

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

#### Explanation:

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

I will start by rewriting the expression:

$5 {\left({x}^{2} + 5\right)}^{4} \left(2 x\right) {\left(x - 3\right)}^{4} + {\left({x}^{2} + 5\right)}^{5} \left(4\right) {\left(x - 3\right)}^{3}$

$= 10 x {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{4} + 4 {\left({x}^{2} + 5\right)}^{5} {\left(x - 3\right)}^{3}$

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

$= 2 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{3} \left[5 x \left(x - 3\right) + 2 \left({x}^{2} + 5\right)\right]$

$= 2 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{3} \left[5 {x}^{2} - 15 x + 2 {x}^{2} + 10\right]$

$= 2 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{3} \left(7 {x}^{2} - 15 x + 10\right)$

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

$\frac{d}{\mathrm{dx}} \left(f g h\right) = f ' g h + f g ' h + f g h '$

(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:
$\frac{d}{\mathrm{dx}} \left(2 f g h\right) = 2 f ' g h + 2 f g ' h + 2 f g h '$

So the derivative of our expression is:

$2 \left[4 {\left({x}^{2} + 5\right)}^{3} \left(2 x\right)\right] {\left(x - 3\right)}^{3} \left(7 {x}^{2} - 15 x + 10\right)$

$+ 2 {\left({x}^{2} + 5\right)}^{4} \left[3 {\left(x - 3\right)}^{2}\right] \left(7 {x}^{2} - 15 x + 10\right)$

$+ 2 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{3} \left[14 x - 15\right]$

We can simplify by first simplifying each term:

$16 x {\left({x}^{2} + 5\right)}^{3} {\left(x - 3\right)}^{3} \left(7 {x}^{2} - 15 x + 10\right)$

$+ 6 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{2} \left(7 {x}^{2} - 15 x + 10\right)$

$+ 2 {\left({x}^{2} + 5\right)}^{4} {\left(x - 3\right)}^{3} \left(14 x - 15\right)$

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
$91 {x}^{4} - 390 {x}^{3} + 690 {x}^{2} - 750 x + 375$

So we end up with:

$2 {\left({x}^{2} + 5\right)}^{3} {\left(x - 3\right)}^{2} \left(91 {x}^{4} - 390 {x}^{3} + 690 {x}^{2} - 750 x + 375\right)$

Notes

1
The derivative of $f g h$ is:

$\frac{d}{\mathrm{dx}} \left(f g h\right) = \frac{d}{\mathrm{dx}} \left(\left(f g\right) h\right) = \left(f g\right) ' h + \left(f g\right) h '$

$= \left(f ' g + f g '\right) h + f g h '$

$= f ' g h + f g ' h + f g h '$

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. :-)