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Differentiate the function y = 3^x

To find the derivative of the function ( y = 3^x ), we apply the chain rule. We start by letting ( u = 3^x ) and thus ( y = f(u) = u ). The chain rule states that:

y=f(u)uy' = f'(u) \cdot u'

Next, we need to determine ( f’(u) ). Since ( f(u) = u ), the derivative is straightforward:

f(u)=ddu(u)=1f'(u) = \frac{d}{du}(u) = 1

Now, we compute ( u’ ), which is the derivative of ( u = 3^x ). Using the formula for the derivative of an exponential function, we have:

u=ddx(3x)=ln(3)3xu' = \frac{d}{dx}(3^x) = \ln(3) \cdot 3^x

Substituting ( f’(u) ) and ( u’ ) back into the chain rule expression gives us:

y=f(u)u=1(ln(3)3x)=ln(3)3xy' = f'(u) \cdot u' = 1 \cdot (\ln(3) \cdot 3^x) = \ln(3) \cdot 3^x

Thus, we conclude that the derivative of ( y = 3^x ) is:

y=3xln(3)y' = 3^x \ln(3)

This result indicates that the slope of the tangent line to the graph of ( y = 3^x ) at any point ( (x, y) ) is given by ( 3^x \ln(3) ). The function ( y = 3^x ) is a prime example of an exponential function, which increases rapidly as ( x ) becomes larger. The derivative ( y’ ) informs us about the rate of growth of the function at any specific point, while the term ( \ln(3) ) represents the growth rate relative to the base of the exponential function.

Answered by: Dr. Sarah Wilson
A-Level Physics Tutor
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