implicit differentiation question, help

find the equation of the tangent to the curve y=2^x + 2^-x at the point (2, 4 1/4).

need help, both me an brother are getting the wrong answer :s-smilie:

thanks, xxx

Reply 1

EierVonSatan

Do you know how to differentiate a^x functions?

If not think about y = a^x => ln y = ln (a^x) = x ln a (so now ln a is a constant) :smile:

Reply 2

joe_ex

i think i do to some extent, i got as far as

dy/dx= 2^x(ln2) + 2^-x(ln2)
substitute 2 into x

= 4ln2 + 1/4ln2
=ln2^4 + ln2^1/4

then i get stuck :s-smilie:

Reply 3

EierVonSatan

4 ln 2 - (1/4) ln 2 = (15/4) ln 2

this is the gradient at the point (2, 17/4) :smile:

Reply 4

Wha'sHisFace

hi...do u happen 2 have the answer to this question...i attempted it...but dont want 2 post answer for it jus to be wrong...if u post the answer i can see whether i got the right thing then post the solution??

Reply 5

joe_ex

yeah sure, the answer is 4y= 15ln2(x-2) + 17

xx

Reply 6

Wha'sHisFace

[mod edit: can you not post full solutions on the forum]

Reply 7

EierVonSatan

Please do not post full solutions :colondollar:

Reply 8

pavrotte

differentiation wise, I managed to solve using the relationship:
2^x=(e^(ln2))^x=e^(xln2)
and similarly for 2^(-x) =>e^(-xln2)
then differentiated each term with the chain rule - thus avoiding implicit differentiation :-)