# If y = 2^x, find dy/dx

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Q: If y = 2^x, find dy/dx.

1) Take Logs of both sides of our equation y = 2^x

So we get: log(y)=log(2^x)

2) Apply relevant log rule to rhs: Log rule: log(a^b) = b . log(a)

[nb: the dot between b and log(a) represents x / multiply / times] :)

So we get: log(y) = x . log(2)

3) Differentiate both sides with respect to x.

LHS: log(y) => (1/y)(dy/dx) [partial differentiation hence we multiply (1/y) by dy/dx]

RHS: x . log(2) => log(2) [log(2) is a constant so x dissapears]

So we get: (1/y)(dy/dx) = log(2)

4) We want to find dy/dx, which is on the LHS. To get this dy/dx on its own we can multiply both sides by y.

So we get: dy/dx = y . log(2)

5) To finish this question we need to sub in for y and then we have an answer for dy/dx.

Recall y=2^x (from our original question)

So we get: dy/dx = (2^x)(log(2)) => our final solution

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