Examples on differentiation using product rule or sum rule
DIFFERENTIATION IS VERY EASY…..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..,…..like for example
y=2x² +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 knowledge……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
- A practical example,
y=2x² +x +1. Find dy/dx
dy/dx = 2×2•X^(2-1) + 1×1• X^(1-1) + 0×1• X^(0-1)
dy/dx= 4•X^(1) + 1• X^(0) + 0• 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)
Please look at this any question please
Before I proceed,
Okay for beginners only, try this and inbox me your answers on the comment box below.… 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 • [4x + 1])
I.e 3x^2 × [4x + 1])
So whenever you observe anything like that, you make use of product rule,
For example, let’s differentiate
- 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
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 × 4 + 4x + 1 × 6x
dy/dx = 3x^2 • 4 + 4x + 1 • 6x
dy/dx = [3×4 x^2 ]+[ 4x × 6x + 1 × 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, the coefficient of x^2, can be added to any coefficient of x^2 ( i.e tomato can add tomato but pepper can’t add with Tomato)
Say this mantra, product rule is very easy five times, then study this again.
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 video on differentiation
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