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.

f(0)=1

f'(x)=5e^(5x)
So, f'(0)=5
Now we have a point and a slope, so the line is

y-1 = 5(x-0)
or
y=5x+1

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3De%5E(5x),y%3D5x%2B1

To find the equation of the tangent line to the function f(x) = e^(5x) at x = 0, we need to know the instantaneous rate of change (also known as the derivative) of the function at that point.

The derivative of f(x) with respect to x can be found by taking the derivative of e^(5x) using the chain rule. The chain rule states that if we have a function f(g(x)), then the derivative is given by f'(g(x)) * g'(x).

In this case, f(g(x)) = e^x and g(x) = 5x. Therefore, f'(g(x)) = e^x and g'(x) = 5.

Now, let's find the derivative of f(x) = e^(5x). Using the chain rule, we have:

f'(x) = f'(g(x)) * g'(x) = e^(5x) * 5 = 5e^(5x)

To find the slope of the tangent line to f(x) at x = 0, we substitute x = 0 into the derivative:

f'(0) = 5e^(5*0) = 5e^0 = 5 * 1 = 5

Therefore, the slope of the tangent line to f(x) at x = 0 is 5.

Now, we need to find the y-intercept of the tangent line. We know that the equation of a straight line is given by y = mx + b, where m is the slope and b is the y-intercept.

Since the tangent line passes through the point (0, f(0)), we can substitute x = 0 into the original function f(x) = e^(5x) to find the y-coordinate:

f(0) = e^(5*0) = e^0 = 1

So the point (0, 1) is on the tangent line.

Now we have the slope (m = 5) and a point (0, 1) on the line. We can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values, we have:

y - 1 = 5(x - 0)
y - 1 = 5x

Finally, we can rearrange the equation to put it in slope-intercept form (y = mx + b):

y = 5x + 1

Therefore, the equation of the tangent line to f(x) = e^(5x) at x = 0 is y = 5x + 1.