derivative of (x)/(x^2 + 5) please show work thanks in advance
are you familiar with the "quotient rule"?
no just beginning calc
what about \[\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}h\] familiar?
thats defination of derivative right
kimmy can you help me pls
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.
would you mind demonstrating that pls? i just started calc and my teacher isnt a good english speaker
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?
u is numerator v is denominator
yes, correct. In this case, what is u and v?
u= x and v= x^2 - 5
what about u' and v'?
u'= 1? v'= 2x?
Correct.. Now plug al these into my "young man" equation.
(1*x^2-5 minus x * 2x)/(x^2 - 5)^2
??
Almost corret, missed one sign
not sure where
-5 isn't it +5
Your v should be x^2+5
right. so whats next?
that's the answer. the first derivative of the function with respect to x.
how come wolfram says different http://www.wolframalpha.com/input/?i=derivative+of+%28x%29%2F%28x%5E2-5%29
Did you spot you typed different function? i see you write x^2-5 at wolfram
crap. i just noticed. its supposed to be x^2 - 5 sorry
then v=x^2-5
huh
Show respect to your teacher :) What's your final answer.
(1*x^2-5 minus x * 2x)/(x^2 - 5) ??
yes, =x^2-5-2x^2/(x^2-5) =-x^2-5/(x^2-5) =then you can proof wolfram's right with this
ahh i see thanks
one last one
derivative of f(x)= (x^4 + x)^3 (x^2 - 1)^2
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.
im confuse as to how to take the derivative with the exponent oustide the paranthesis
Ex: d(u)^5/dx = 5*(u)^(5-1) * d(u)/dx
so 3 * (4x) (2x) * 2?
Nope, what's u'?
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