i cant seem to differentiate this..?

Question

y = x(2^x)...wad's the dy/dx for this?..answer given is 2^x ( 1 + xln 2)...

Answer 1

y = x(2^x)

Take ln of both sides:

ln y = ln x + x ln 2

Implicit differentiation:


(1/y) dy = (1/x) dx + (ln 2) dx


Multiply both sides by y ( or x(2^x) ):

dy/dx = x(2^x)/x + x(2^x) ln 2


Simplify:

dy/dx = (2^x) + x(2^x) ln 2

distributive property:

dy/dx = 2^x (1 + x ln 2)

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Second method (the product rule):

y = x (2 ^ x)

u = x
du = dx
v = 2 ^ x
dv = (2^x) ln 2

dy/dx = x (2 ^ x) ln 2 + (2 ^ x)

distributive law and switching the terms:
:

dy/dx = 2^x ( 1 + x ln 2)

Answer 2

With any power where x is in the index it is usually necessary to change to a power of e first. This is done by saying let 2^x = e^y and then take logs (base e) of both sides to find y. This is why a ln appears in the answer.

Answer 3

2^x = (e^1n2)^x = e ^ (x ln2)

so (d/dx) 2^x = ( e^ xln2 ) ln 2 = ln 2 * 2^x, by the chain rule. Apply the product rule, and you should be set.

Answer 4

First, change 2^x to e^(ln(2) * x)

x * e^(ln(2) * x)

Now just use the Product Rule.

Answer 5

take log on both the sides & use binomial theorem formulas

Answer 6

Yeah, that looks 'bout right to me.

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


Doug