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)#