What is the derivative of y=2^x?
ln y = x ln 2
1/y dy/dx = ln 2
dy/dx = y ln 2
or
dy/dx = 2^x ln 2
change y = 2^x
to
ln y = x ln2
now take derivative:
dy/dx / y = ln 2
dy/dx = y ln2
= ln2 (2^x)
y'=2^x*ln2
if y=a^x
y'=a^x *lna
Reiny's solution is cool, using implicit differentiation. You can also do it by noting that
2 = e^(ln 2)
So,
2^x = e^(ln 2 * x)
and now use the chain rule:
d/dx e^u = e^u du/dx, so
d/dx e^(ln 2 * x) = ln2 * e^(ln 2 * x) = ln2 * 2^x
To find the derivative of the function y = 2^x, we can make use of logarithmic differentiation. Here's how you can do it step by step:
Step 1: Take the natural logarithm (ln) of both sides of the equation: ln(y) = ln(2^x).
Step 2: Apply the logarithmic property of exponents: ln(y) = x * ln(2).
Step 3: Differentiate both sides of the equation with respect to x. The derivative of ln(y) with respect to x is (1/y) * dy/dx. The derivative of x * ln(2) with respect to x is simply ln(2).
Step 4: Replace ln(y) with y and rearrange the equation to solve for dy/dx: (1/y) * dy/dx = ln(2).
Step 5: Multiply both sides of the equation by y: dy/dx = y * ln(2).
Step 6: Substitute y with 2^x: dy/dx = 2^x * ln(2).
Therefore, the derivative of y = 2^x is dy/dx = 2^x * ln(2).