Question

What is the equation of the line tangent to `f(x)= x^5-e^x` at `x=2`?

Answer 1

`y=(80-e^2)*x + e^2-128`

exactly or

`y=72.61 x-120.61`

approximately.

Explanation:

You need to take the derivative and the evaluate it at x=2.

We have

`f(x)=x^5 -e^x`.

We take

`{d f(x)}/dx = {d }/dx ( x^5 - e^x) = {d }/dx ( x^5) - {d }/dx ( e^x)`

using the Linearity of differentiation.

Now we apply the rules for taking the derivative of powers and exponential functions, `D_x x^n = n x^{n-1}` and `D_x e^x =e^x` (isn't Euler's notion so compact?).

`{d f(x)}/dx = 5*x^4 - e^x`

We need to evaluate this at `x=2`, because this is the slope of the tangent line at x=2.

`m=5*2^4 - e^2=80-e^2~= 72.61`.

To find the equation of this line we need to know a point on the line, the only point we know is that the line touches the function at
`x=2`, so `(2,(f(2))`, is on the line.

We need `f(2)`,

`f(2)=2^5 -e^2=32-e^2~=24.61`.

Recall the equation of the line,

`y=m*x+b`,

so we have

`32-e^2 = (80-e^2)*2+b=160-2*e^2+b`.

Rearranging we get

`b=32-e^2-160+2e^2=e^2-128~=-120.61`.

Putting everything back into the line we have,

`y=(80-e^2)*x + e^2-128`

exactly or

`y=72.61 x-120.61`

approximately.