How do you do the derivative of a^x, for example, (3/4)^x
Thank you for helping!
recall that if y = e^u where u is a function of x,
y' = e^u du/dx
Now, a^x = (e^(ln a))^x = e^(x*ln a)
so, y' = e^(x*ln a) * ln a = ln a * a^x
makes sense, since if a>e, ln a > 1, and the curve rises more steeply
Use the exponent rule. Bring the exponent down in front and then reduce your exponent by 1.
a^x = x(a^(x-1))
(3/4)^5 = 5(3/4)^4
To find the derivative of a function of the form a^x, where a is a constant, you can use logarithmic differentiation.
Here's how you can find the derivative of (3/4)^x:
Step 1: Take the natural logarithm (ln) of both sides of the equation:
ln(y) = ln((3/4)^x)
Step 2: Use the logarithmic property that ln(a^b) = b ln(a):
ln(y) = x ln(3/4)
Step 3: Differentiate both sides of the equation with respect to x:
(d/dx) ln(y) = (d/dx) (x ln(3/4))
Step 4: Apply the chain rule on the right side of the equation:
(d/dx) ln(y) = (d/dx) x * ln(3/4) + x * (d/dx) ln(3/4)
The derivative of ln(y) with respect to x is simply (1/y) * (dy/dx), and (d/dx) x is just 1:
(1/y) * (dy/dx) = 1 * ln(3/4) + x * (d/dx) ln(3/4)
Step 5: Solve for (dy/dx) or y':
(dy/dx) = y * [ln(3/4) + x * (d/dx) ln(3/4)]
Step 6: Remember that y is the original function (3/4)^x:
(dy/dx) = (3/4)^x * [ln(3/4) + x * (d/dx) ln(3/4)]
So, the derivative of (3/4)^x with respect to x is (3/4)^x multiplied by the derivative of ln(3/4) plus x times the derivative of ln(3/4).
Note: The derivative of ln(3/4) is simply the derivative of ln(3) minus the derivative of ln(4). The derivatives of ln(3) and ln(4) are constants, as ln(x) has a derivative of 1/x.