Differentiation using quotient rule: Can somebody please tell me how to solve this red arrow part, see the attached image file

8-12=-4, so you have -4x^2 That part is just algebra

No, I meant to say where we got that 8^4 +24x part

8x^3*x from previous step

yes but how, can you please explain it ..

(x * y') - (x' * y) / (x)^2

Simplify this \[ \frac{(x^3 + 3)8x - 4x^2 (3x^2)}{(x^3+2)^2}\]

(3x^3+8x-4x^2) / (x^3+2)^2

oh sorry, I didn't see the power signs are too small .. (x^3+3 +8x) - (12x^2) / (x^3+3)^2

When you simplify it you will get the final answer shown below the red arrow.

yes, I am not good at simplifying. I almost forgot it.

your steps are incorrect!! how can you be doing calculus ... ?? i think you should start with basic algebra

Let try basic algebra : (x³+3) 8x = ...

can you please mention a link or tutorial where I can learn simplification and algebra. I don't think I am able to do derivative unless I learn some basic algebra.

Can distribute ( multiply) 8x into the parentheses ?

you?

8x^3+3 ?

How about: 8x * x³ = ..

24x^3 ?

Add exponents when you multiply

or it could be 8x and add powers? = 8x^4 ?

Don't get confused between exponent and the constant:

Yep, 8x^4 ( sigh !)

3* 8x = ?

24x , right?

Yep, now (x³+3) 8x = ...

8x^4 + 24x

and in the last part .. 4*3 = 12 ... then add exponents 12x^4

Great :) 4x² * 3x² =

Good job, now put them together!

Thank you so much @Chlorophyll .. Can you please give me a tutorial link where I can do exercises of simplifcations?

No more confused!

thank you very much all of you. :) this community is really helpful.

http://www.regentsprep.org/regents/math/ALGEBRA/AO5/PracExp.htm

Hope you'll improve greatly at your next post :)

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