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Mathematics 19 Online
OpenStudy (anonymous):

derivative of (x)/(x^2 + 5) please show work thanks in advance

OpenStudy (lgbasallote):

are you familiar with the "quotient rule"?

OpenStudy (anonymous):

no just beginning calc

OpenStudy (lgbasallote):

what about \[\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}h\] familiar?

OpenStudy (anonymous):

thats defination of derivative right

OpenStudy (anonymous):

kimmy can you help me pls

OpenStudy (anonymous):

Use this young man: \[\frac{d}{dx}(\frac{u}{v})=\frac{u'v-uv'}{v^2}\] Where u' and v' means du/dx and dv/dx.

OpenStudy (anonymous):

would you mind demonstrating that pls? i just started calc and my teacher isnt a good english speaker

OpenStudy (anonymous):

I can't, i'm not in my mood to work something myself. But I can help you. Can you show me what is u and v in your problem?

OpenStudy (anonymous):

u is numerator v is denominator

OpenStudy (anonymous):

yes, correct. In this case, what is u and v?

OpenStudy (anonymous):

u= x and v= x^2 - 5

OpenStudy (anonymous):

what about u' and v'?

OpenStudy (anonymous):

u'= 1? v'= 2x?

OpenStudy (anonymous):

Correct.. Now plug al these into my "young man" equation.

OpenStudy (anonymous):

(1*x^2-5 minus x * 2x)/(x^2 - 5)^2

OpenStudy (anonymous):

??

OpenStudy (anonymous):

Almost corret, missed one sign

OpenStudy (anonymous):

not sure where

OpenStudy (anonymous):

-5 isn't it +5

OpenStudy (anonymous):

Your v should be x^2+5

OpenStudy (anonymous):

right. so whats next?

OpenStudy (anonymous):

that's the answer. the first derivative of the function with respect to x.

OpenStudy (anonymous):

how come wolfram says different http://www.wolframalpha.com/input/?i=derivative+of+%28x%29%2F%28x%5E2-5%29

OpenStudy (anonymous):

Did you spot you typed different function? i see you write x^2-5 at wolfram

OpenStudy (anonymous):

crap. i just noticed. its supposed to be x^2 - 5 sorry

OpenStudy (anonymous):

then v=x^2-5

OpenStudy (anonymous):

huh

OpenStudy (anonymous):

Show respect to your teacher :) What's your final answer.

OpenStudy (anonymous):

(1*x^2-5 minus x * 2x)/(x^2 - 5) ??

OpenStudy (anonymous):

yes, =x^2-5-2x^2/(x^2-5) =-x^2-5/(x^2-5) =then you can proof wolfram's right with this

OpenStudy (anonymous):

ahh i see thanks

OpenStudy (anonymous):

one last one

OpenStudy (anonymous):

derivative of f(x)= (x^4 + x)^3 (x^2 - 1)^2

OpenStudy (anonymous):

d(u*v)/dx = u'v+uv' ..........................(2) You know the routine: determine u and v, find u' and v' plug all these into (2). Show me your final answer.

OpenStudy (anonymous):

im confuse as to how to take the derivative with the exponent oustide the paranthesis

OpenStudy (anonymous):

Ex: d(u)^5/dx = 5*(u)^(5-1) * d(u)/dx

OpenStudy (anonymous):

so 3 * (4x) (2x) * 2?

OpenStudy (anonymous):

Nope, what's u'?

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