## Q:I'm having trouble with taking the derivative of functions. Could you help me by taking the derivative of this function please? "f(x) = (2-3x^2)^5/5x"

Ah, derivatives. I guess you’re getting into Calculus, huh?

Since you’re in Calculus, I’m going to assume you’re smart enough to know how a lot of the math rules work, like order of operations and basic algebra and such. So I’m going to go straight into the calculus part.

Calculus can be a harder concept to grasp, but it’s necessary to solve the bigger questions in Maths. I’ve never taken any higher maths than Calculus, so I couldn’t say for sure what it’s used for exactly, but I assure you, Calculus is necessary and a huge deal if you plan on pursuing maths further. In all honesty, 99.999% of people will need no maths further than basic algebra, and maybe simple physics, but that’s another discussion all together.

Personally, I love math and science. I love solving problems, and I really considered being a Mathematician instead of IT. I just really like working with numbers and solving life’s greatest riddles, in terms of numbers that is.

Anyways, enough with my rambling, you probably want the answer!

Let’s examine what we got here…

f(x) = (2-3x^2)^5/5x

It’s going to be hard to format this in an easily readable way on Tumblr, so I hope this is legible. I would show my work on paper and upload a picture, but I have no scrap paper handy at the moment.

We got a 2 different rules going on here. The Power Rule and the Quotient Rule. The Power Rule is more simpler to grasp in my mind. The Quotient Rule is usually the one that trips people up.

We can do both the Quotient Rule and the Power Rule at the same time!

Let’s examine the Quotient Rule:

d/dx (u/v) = (u’v-uv’)/v^2

I am also going to assume you know the tics next to the letters mean “prime,” which means the derivative of.

What we do here is that, on top, we put together the derivative of the top part with the bottom part and we take the difference with the top part and the derivative of the bottom part, all over the bottom part squared, if that makes sense

For the Quotient Rule (and also the more complex rules like the Chain Rule and the Product Rule), I find it’s easier to break down both halves.

So, we have…

u = (2-3x^2)^5, or the Top Part

v = 5x, or the Bottom Part

We are going to need to take the derivative of both halves, so let’s do that now.

v’ should be easy, as the power of the x is one.

d/dx (5x) = 5

v’ = 5

No thrills there.

The top part is a little more interesting, as we got the Power Rule here. We did technically use the Power Rule for calculating v’, but here, it’s easier to see.

The Power Rule is described as…

d/dx (u^n) = nu^(n-1)(du/dx)

What this means is that we drop down the power in front of the function we are using the power rule for. We subtract one from the standing power when we do this, which is why 5x went to 5. x was to the first ;power, which when dropped down multiplied with 5, which is still 5. We subtract one from the power, which makes it x to the 0th, which simplifies down to 1, so we are left with 5. Now, the other part comes into play with more complex stuff, such as we got here. The (du/dx) part is the derivative of the original u function. We just stick that with what we did and we got a derivative.

Let’s do this then.

d/dx ((2-3x^2)^5) = 5(2-3x^2)^4(-6x) = -30x(2-3x^2)^4

u’ = -30x(2-3x^2)^4

So what we did is drop the 5 down in front and reduced the power to 4, then did the derivative of the inside. The derivative of any constant is 0, so we don’t need to worry about the 2. -3x^2 is another Power Rule in itself. We drop the 2 down and reduce that power to 1 to make -6x. I then simplified and combined the 5 and the -6x to get -30x, and slapped the rest at the end.

Now that we got all 4 components, we can slap them in the Quotient Rule formula.

d/dx ((2-3x^2)^5/5x) = ((-30x(2-3x^2)^4)(5x)-((2-3x^2)^5)(5))/(5x)^2

No we just need to simplify and combine everything

d/dx ((2-3x^2)^5/5x) = (((-150x^2)(2-3x^2)^4)-(5(2-3x^2)^5))/(25x^2)

So, the final answer is…

(((-150x^2)(2-3x^2)^4)-(5(2-3x^2)^5))/(25x^2)

Sorry for all the parentheses, I tried to make it look as clean as I could in Tumblr.

I hope that helps, and good luck with the Calculus!

I can also do Integrals too if you need help with those!

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