Find dy/dex: a^xy + ln(xy) = xy?

Question

been awhile since I've done this and the book did a pretty bad job of teaching it; have to show all my work =/

Answer 1

let u = xy
then du/dx = y
let a^u = e^z
then z = ulna
a^u = e^ulna
a^u + lnu = e^ulna + lnu
d(e^ulna + lnu)/du = (lna)e^ulna + 1/u
d(a^xy + ln(xy))dx = y(ln(a))a^xy + y/xy =
(xy(ln(a))a^xy + 1)/x let u = xy
then du/dx = y
let a^u = e^z
then z = ulna
a^u = e^ulna
a^u + lnu = e^ulna + lnu
d(e^ulna + lnu)/du = (lna)e^u + 1/u
d(a^xy + ln(xy))dx = y(ln(a))e^xy + y/xy =
(xy(ln(a))a^xy + 1)/x ≠ xy

Answer 2

Before we get started, let's list our derivatives.

d/dx e^x = e^x
d/dx a^x = a^x (ln a)

Therefore, for

a^(xy) + ln(xy) = xy

Differentiating implicitly, we get

a^(xy) ln(a) [y + (1)dy/dx] + 1/(xy) * [y + dy/dx] = [y + dy/dx]

Expanding everything, we get

a^(xy) ln(a) y + a^(xy) ln(a) [dy/dx] + 1/x + 1/(xy) [dy/dx] = y + dy/dx

Everything with a dy/dx in it goes to the left hand side; everything else, to the right hand side.

a^(xy) ln(a) [dy/dx] + 1/(xy) [dy/dx] - [dy/dx] = -a^(xy) ln(a) y - 1/x + y

Factoring out dy/dx,

[dy/dx] (a^(xy) ln(a) + 1/(xy) - 1) = -a^(xy) ln(a) y - (1/x) + y

Now, we just have to divide to isolate the dy/dx

dy/dx = [-a^(xy) ln(a) y - (1/x) + y] / (a^(xy) ln(a) + 1/(xy) - 1)

We can, in theory, simplify the problem more (since there are fractions within that fraction), but I'll leave it at that. You'd specifically multiply top and bottom by xy to get rid of all the fractions. You know, why don't we just do that; I have time.

dy/dx = [-a^(xy) ln(a) xy^2 - y + xy^2] / (a^(xy) ln(a)xy + 1 - xy)

Answer 3

few things to remember first:
dy/dx of a^ f(x) = (ln a) [a^ f(x)] f'(x)

dy/dx of ln [f(x)] = f'(x)/f(x)

dy/dx of xy: dy/dx of x + derivative of y * dy/dx = 1 + dy/dx then rearrange equation for dy/dx

now d[a^xy + ln(xy) = xy]/dx:

ln a [a^xy] [ 1 + dy/dx] + [1 + dy/dx]/xy = 1 + dy/dx

ln a (a^xy) + ln a (a^xy) * dy/dx + 1/xy + dy/dx * 1/xy = 1 + dy/dx

ln a (a^xy) * dy/dx + dy/dx * 1/xy - dy/dx = 1 - ln a (a^xy)

dy/dx * [ln a (a^xy) + 1/xy - 1] = 1 - ln a (a^xy)

dy/dx = [1 - ln a (a^xy)] / [ln a (a^xy) + 1/xy - 1]

hope that makes sense to you =)

Answer 4

Assume z = xy, our eqn becomes

a^z + ln(z) = z

differentiate both sides wrt 'x'

[a^z*ln(a) + (1/z)]*(dz/dx) = (dz/dx)

[ a^(xy)*ln(a) + 1/(xy) ]*(y + xy') = y + xy'

y*a^(xy)*ln(a) + (1/x) + [ x*a^(xy)*ln(a) + (1/y) ]*y' = y + xy'

y*{ a^(xy)*ln(a) - 1 } + (1/x)

+

[ {x*a^(xy)*ln(a) - 1 } + (1/y) ]*y' = 0

y' * [ {x*a^(xy)*ln(a) - 1 } + (1/y) ] = - [ y*{ a^(xy)*ln(a) - 1 } + (1/x) ]

y' = - [ y*{ a^(xy)*ln(a) - 1 } + (1/x) ] / [ {x*a^(xy)*ln(a) - 1 } + (1/y) ]

Answer 5

Isn't this dy/d(e^x) , or just a typo?

If it is, then you can follow the following hint:

dy/d(e^x) = (dy/dx)[dx/d(e^x)] = (dy/dx)/e^x