# Easy method of learning differentiation

## Examples on differentiation using product rule or sum rule

DIFFERENTIATION IS VERY EASYâ€¦..I want you to repeat that mantra three times, for those that have develop this mindset of, CALCULUS IS HARD

Now let’s look at simple differentiation..,â€¦..like for example

y=2xÂ² +x +1. Find dy/dx

The power of the first x is 2, so the coefficient of the same first x is 2

For the second x

The power is 1 and the coefficient is still 1

And the last, has just 1( and invisible x^0)
So after identifying this knowledgeâ€¦â€¦you can proceed to the next step

Differentiation

dy/dx = (power of x) multiply by (the coefficient of x) and remove 1 from the power of x

1. A practical example,

### y=2xÂ² +x +1. Find dy/dx

dy/dx = 2Ã—2â€¢X^(2-1) + 1Ã—1â€¢ X^(1-1) + 0Ã—1â€¢ X^(0-1)

dy/dx= 4â€¢X^(1) + 1â€¢ X^(0) + 0â€¢ X^(-1)

dy/dx= 4X^1 + 1 + 0

dy/dx= 4X+ 1 ( final answer)

Note: anything raise to power zero (0) is One (1)

And zero(0) multiply by anything is zero (0)

Before I proceed,

Okay for beginners only, try this and inbox me your answers on the comment box below.â€¦ I’ll reveal the winner

### PRODUCT RULE

Okay next topic is on product rule, we just finished what we know as sum rule
Product rule is very simple easy and interesting say that five times, so it can stick (with God all things are possible)

What’s product rule? Product simply means the multiple of two or more things.

For example 3x^2 [4x + 1]

How do you differentiate? But you’ll observe that the both of them are connecting with product rule

For instance, (3x^2 [4x + 1]) = 3x^2 â€¢ [4x + 1])

I.e 3x^2 Ã— [4x + 1])

So whenever you observe anything like that, you make use of product rule,

For example, let’s differentiate

Â

• y = 3x^2 [4x + 1]

Solution
y = 3x^2 [4x + 1]
We can’t differentiate this directly, then we’ll apply product rule.

Product rule, dy/dx = Udv + vdu

Â What’s U and what’s V?

In . y = 3x^2 [4x + 1]
Let u = 3x^2
And V = 4x + 1

Now differentiate U and V (using sum rule)

Let u = 3x^2
du/dx = 6x

And V = 4x + 1
dv/dx = 4

Recall from product rule
dy/dx = Udv/dx + vdu/dx

So substitute the value for u, dv/dx, v and du/dx there

Product rule, dy/dx = Udv + vdu

dy/dx = 3x^2 Ã— 4 + 4x + 1 Ã— 6x
dy/dx = 3x^2 â€¢ 4 + 4x + 1 â€¢ 6x
dy/dx = [3Ã—4 x^2 ]+[ 4x Ã— 6x + 1 Ã— 6x]

dy/dx = [12x^2 ]+[ 24x^2 + 6x]
=12x^2 +24x^2 + 6x
=36x^2 + 6x ( final answer)

Note: alike terms can add themselves but unlike terms can’t. For instance,Â the coefficient of x^2, can be added to any coefficient of x^2 ( i.e tomato can add tomato but pepperÂ  can’t add with Tomato)

Say this mantra, product rule is very easy five times, then study this again.Â

Related content

Do you know limit very well, if not check this now (limit@infinity)

Find the limits of 7-3x^2+6x^5/ 2+x-3x^5 as x approaches infinity

In other words, find the limits of 7- 3x square + 6x raise to power 5/ 2 + x – 3x raise to power 5

Watch videoÂ on differentiationÂ

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YOU CAN DO IT

### Oluwamuyide Peter

On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide ðŸŒŽ

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