# determine the equation of the line tangent to the graph of `y=14e^(9x)` ` `

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### 2 Answers

Equation of a tangent line at point `(x_0,y_0)` to the graph of `y=f(x)` is

`y-y_0=f'(x_0)(x-x_0)`

So let's calculate `f'(x)`

`(14e^(9x))'=14e^(9x)cdot9=126e^(9x)`

Hence equation of the tangent line at the point `(x_0,y_0)` is

`y-y_0=126e^(9x_0)(x-x_0)`

For example equation of the tangent line at point (0,14) would be

`y-14=126e^(9cdot0)(x-0),`

`y=126x+14`

Blue is your function `y=14e^(9x)` and red is the tangent line at point (0,14).

### User Comments

`y=14e^(9x)`

`m=y'=126e^(9x)`

`y-y_0=126e^(9x) (x-x_0)`

`y=126x e^(9x_0)-126x_0 e^(9x_0)+y_0`