How do you find the equation of the line tangent to #y=sqrtx# at (1,1)?

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Answer 1 · 2016-12-09T04:10:53

#y=1/2x+1/2#

The slope of the tangent line to a function #f# at the point #x_1# is equal to the value of the derivative at that point, or #f'(x_1)#.

Another notation states that the tangent line to some function #y# at #x_1# is equal to #dy/dx|_(x=x_1)#.

For #f(x)=sqrtx#, we will be able to differentiate this using the power rule if we first rewrite the square root.

#f(x)=sqrtx=x^(1/2)#

The power rule states that the derivative of #x^n# is #nx^(n-1)#.

#f'(x)=1/2x^(1/2-1)=1/2x^(-1/2)=1/(2x^(1/2))=1/(2sqrtx)#

We are interested in the point #(1,1)#, which is at #x=1#. So the slope of the tangent line at this point is:

#f'(1)=1/(2sqrt1)=1/2#

We can relate a slope of #1/2# and the point #(1,1)# using the point-slope equation:

#y-y_1=m(x-x_1)#

#y-1=1/2(x-1)#

#y=1/2x+1/2#

Both the original function and tangent line graphed:

graph{(y-sqrtx)(y-(1/2x+1/2))=0 [-0.893, 6.902, -0.28, 3.617]}