Question

Find the slope of the line tangent to p(x)=(2^x)/(x^2+1) at x=3

Answer 1

\[p(x)=\frac{ 2^x }{ x^2+1 }\] is the equatio from teh problem

Answer 2

find the 1st derivative and substitute x = 3 to find the slope.

Answer 3

what is the 1st derivative I can't figure it out

Answer 4

Quotient rule with a bit of a trick to it. The derivative of any constant c^x is c^x * ln(c). Something nice to remember and we will have to use it here.
So let's review the quotient rule:
\[\frac{ g'(x)h(x) - g(x)h'(x)}{ h(x)^2}\]
Where g(x) is the function in the numerator, h(x) the function in the denominator.
So apply this to our function:
The derivative of the top:
\[\frac{ d }{ dx } (2^x) = 2^x * \ln(2)\]
Derivative of the bottom:
\[\frac{ d }{ dx } (x^2 + 1) = 2x\] (Constants equal zero)
Plug it all in, square the denominator:
\[\frac{ (2^x \ln(x))(x^2 + 1) - 2^x * 2x }{ (x^2 + 1)^2 }\]
Simplify if you please. If I can also get someone to check my work that would be wonderful.

Answer 5

oops you final should read

(\[\frac{dy}{dx} = \frac{2^xlog(2)\times(x^2 +1) - 2^x \times 2x}{(x^2 + 1)^2}\]

Answer 6

Where was my mistake?

Answer 7

the final answer is .7044 correct?

Answer 8

you have \[2^xlog(x).... \]

and shouldn't it be

\[2^x \log(2)\]

Answer 9

is my answer correct?

Answer 10

Oh yes, didn't catch that. Thanks for the correction. And hold on farmergirl I'm checking it

Answer 11

ok

Answer 12

here is the wolfram alpha version...

http://www.wolframalpha.com/input/?i=differentiate+y+%3D+%282%5Ex%29%2F%28x%5E2+%2B1%29

Answer 13

This derivative is correct according to Wolfram Alpha:
\[ \frac{(2^x*\ln(2))(x^2+1) − 2^x*2x }{ (x^2+1)^2 }\]
And no farmergirl, your answer is incorrect. This is a bit of a tedious derivative to computate so if you give up, where's the link to the answer
:
https://www.wolframalpha.com/input/?i=%28%282^x*ln%282%29%29%28x^2%2B1%29+%E2%88%92+2^x*2x%29%2F%28x^2%2B1%29^2+%2C+x+%3D+3

Answer 14

Just click on Approximate Form

Answer 15

so then is the answer 100 or was my other answer correct

Answer 16

Wolfram Alpha is telling me that the answer is appox. 0.0745

Answer 17

ok that is the answer that I got before

Answer 18

p(x)=P(x)=2^x/(x^2+1)
Diff. w .r.t. xType equation here.
P’(x)=((x^2+1) 2^x log2-2^x*2x)/((〖x^2+1)〗^2 )
At x=3
P’(x)=((3^2+1)2^3 log2-2^3*2*3))/(3^2+1)^2
P’(x)=(10*8log2-8*6)/100
P’(x)=.8log2-.48