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

1 Answer
Jan 25, 2016

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)