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# 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?

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#### Explanation

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#### Explanation:

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2
Jun 18, 2018

$y = 5 x + 1$

#### Explanation:

The rate of change function of $f \left(x\right)$ is its first derivative $f ' \left(x\right) = 5 {e}^{5 x}$. At $x = 0$, $f ' \left(0\right) = 5$.

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

To determine $c$, we make the values of the two curves match at the same point, $x = 0$. $f \left(0\right) = 1$, so $1 = 5 \cdot 0 + c \Rightarrow c = 1$.

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

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]}

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