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Imagine a curve where at every single point, the slope of the graph equals the exact height of the graph. That unique curve is the natural exponential function, y=exy = e^x.
In the 1720s, Swiss mathematician Leonhard Euler formalized the base constant e≈2.71828e \approx 2.71828, discovering that the derivative of exe^x is simply itself: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x.
What happens when the exponent is scaled by a constant factor, like e3xe^{3x} or e−0.5xe^{-0.5x}?

The Derivative Rule for e^(kx)

When a constant kk multiplies xx in the exponent, we apply the chain rule — a calculus method where we differentiate the outer function and multiply by the derivative of the inner expression.
Treating u=kxu = kx as the inner function, ddx[u]=k\frac{d}{dx}[u] = k. Multiplying the outer derivative ekxe^{kx} by kk gives the general differentiation rule: ddx[ekx]=kekx\frac{d}{dx}\left[e^{kx}\right] = k e^{kx}
How do we use this derivative to find the exact line tangent to an exponential curve?

Finding Slopes and Tangent Lines

A tangent line is a straight line that touches a curve at a single point, sharing the curve's exact instantaneous slope at that location.
Let's find the tangent line equation for f(x)=4e2xf(x) = 4e^{2x} at the point where x=0x = 0. First, find the yy-value: f(0)=4e2(0)=4(1)=4f(0) = 4e^{2(0)} = 4(1) = 4, giving the contact point (0,4)(0, 4).
Next, compute the derivative: f′(x)=4⋅(2e2x)=8e2xf'(x) = 4 \cdot (2e^{2x}) = 8e^{2x}. At x=0x = 0, the slope is m=f′(0)=8e0=8m = f'(0) = 8e^{0} = 8.