Question

For `f(x)=2^x` what is the equation of the tangent line at `x=1`?

Answer 1

Given that we have to find the tangent of the line `f(x)=2^x` at `x=1`.
From general slope equation, we know that `y-y_o=m(x-x_o)`
Also, we also know that the slope of a tangent at a point of the line is the derivative of the line at that point, i.e `frac{dy}{dx}|_(x=x_o)=m`

So, `\frac{dy}{dx}=\frac{d}{dx}(2^x)=2^xlog2`
Subsequently, at `x=1` `m=2^1log2=2log2`

Now, to find `x_o` and `y_o`
From the given condition, we know that `x_o=1` (because that's where we want to take the tangent from on the x-axis), so to find `y_o` we just place the value of `x_o` into the function. So `y_o=2^(x_o)=2^1=2`
So `x_o=1` and `y_o=2`

Taking these values of `m`, `y_o` and `x_o` into account and substituting these values into the general slope equation, we get
`y-2=2log_e(2)(x-1)`