Question

Use the instantaneous rate of change of `f(x) = e^(5x)` to find the equation of the tangent line to f(x) at x = 0?

Answer 1

`y=5x+1`

Explanation:

The rate of change function of `f(x)` is its first derivative `f'(x)=5e^(5x)`. At `x=0`, `f'(0)=5`.

This is also the gradient of the line tangent to the curve, `y=mx+c`. So `m=5`.

To determine `c`, we make the values of the two curves match at the same point, `x=0`. `f(0)=1`, so `1=5*0+crArrc=1`.

Thus the equation of the tangent line at `x=0` is `y=5x+1`.

Sanity check the answer by plotting the two curves on the same graph and observing that they touch in the right place:
graph{(y-(e^(5x)))(y-(5x+1))=0 [-1, 1, -0.206, 2.291]}