Problem : Use the chain rule method and plss show ur solution so i can understand it
y=(x^2+5)^5/(x^4+3)^3

Question

Problem : Use the chain rule method and plss show ur solution so i can understand it
y=(x^2+5)^5/(x^4+3)^3

anonymous · Answer

WAT DO U WANT

calculusfunctions · Answer

You would like to differentiate y w.r.t. x, correct?

calculusfunctions · Answer

You can use the chain rule within the quotient rule.

anonymous · Answer

$$y'=(x ^{2}+5)^{5}\div(x ^{4}+3)^{3}$$

anonymous · Answer

im confuse.

calculusfunctions · Answer

The given function is the derivative function? So the first time, you typed it incorrectly? There seem to be two options here. Either find the derivative or the antiderivative (integral)? Which is it? No one can help you unless you make your intentions clear by typing the question properly.

anonymous · Answer

derivative sir

calculusfunctions · Answer

You mean you're given$$\frac{ dy }{ dx }=\frac{ (x ^{2}+5)^{5} }{ (x ^{4}+3)^{3} }$$and you would like to find the anti-derivative. Is that correct?

anonymous · Answer

the derivative

calculusfunctions · Answer

So you mean it's$$y =\frac{ (x ^{2}+5)^{5} }{ (x ^{4}+3)^{3} }$$and you would like to find the derivative? Which is it?

anonymous · Answer

using chain rule . find derivative y'

calculusfunctions · Answer

OK then finally we're getting somewhere! Do you know the quotient rule?

Quotient Rule:

If  y = f/g  then$$y \prime =\frac{ f \prime g -fg \prime }{ g ^{2} }$$Do you know this rule?

anonymous · Answer

i know the method but im really confused

calculusfunctions · Answer

About what?

anonymous · Answer

all this problem

calculusfunctions · Answer

OK we have$$y =\frac{ (x ^{2}+5)^{5} }{ (x ^{4}+3)^{3} }$$Let's take baby steps and start by letting$$f =(x ^{2}+5)^{5}$$and$$g =(x ^{4}+3)^{3}$$Now do you know how to find f' and g'?

anonymous · Answer

$$f=4(x ^{2}+5)^{4}(2x)$$

calculusfunctions · Answer

No. Try again!

anonymous · Answer

$$f=5(x ^{2}+5)^{4}(2x)$$

calculusfunctions · Answer

Yes you can!! See! Very good except that it's f'. You wrote f.

calculusfunctions · Answer

Now find g' similarly.

anonymous · Answer

is (2x) have exponent of ^5 ??

calculusfunctions · Answer

No.

anonymous · Answer

$$g=3(x ^{4}+3)^{2}(4x ^{3})$$

anonymous · Answer

ahh ok

calculusfunctions · Answer

Yes and you mean g'. Why do you keep refusing to indicate the prime for derivative? Without that one simple symbol, it's wrong. Understand?

anonymous · Answer

ah sorry im just lazy typing that :)

calculusfunctions · Answer

OK now substitute f, f', g, and g' into the quotient rule formula I posted above.

calculusfunctions · Answer

Well, as a teacher I'm telling you it's wrong without that symbol and as a teacher it's really irritating to see you do that time and time again, even after I tell you not to. I would give you a zero just for that.

anonymous · Answer

$$y'=\frac{ 5(x ^{2}+5)^{4}(2x) }{ 3(x ^{4}+3)^{2}(4x ^{3}) }$$

calculusfunctions · Answer

NO! That's not the formula I gave you! Stop being so lazy and look!!

anonymous · Answer

ahh i know it

anonymous · Answer

pellet im still confused

calculusfunctions · Answer

$$f=(x ^{2}+5)^{5}$$$$g =(x ^{4}+3)^{3}$$$$f \prime =5(x ^{2}+5)^{4}(2x)$$$$g \prime =3(x ^{4}+3)^{2}(4x ^{3})$$Now$$y \prime =\frac{ f \prime g -fg \prime }{ g ^{2} }$$You're telling me you're confused about plugging things into a formula?? Really??

anonymous · Answer

ok i should just solve this using this formula

calculusfunctions · Answer

Not solve, SIMPLIFY!!

anonymous · Answer

ok just calm down .. huhuhu in just a noob

anonymous · Answer

*im

calculusfunctions · Answer

Don't expand the denominator though, only simply the numerator, preferably by factoring.

calculusfunctions · Answer

If properly expressed, the numerator should be in factored form in the final step. Good luck! I have to go now.

anonymous · Answer

tnx