How do you find the equation of a line tangent to #y=sqrt(x+1)# at (3,2)?

Question

2115 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T03:00:55

#y=1/4x+5/4#

The slope of the tangent line will be equal to the value of the derivative evaluated at #x=3#.

#y=(x+1)^(1/2)#

The chain rule states that the derivative of some function to the #1//2# power is:

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

Which comes from the power rule. Thus:

#dy/dx=1/2(x+1)^(-1/2)d/dx(x+1)#

The derivative of #x+1# is just #1#, so the derivative simplifies to:

#dy/dx=1/(2sqrt(x+1))#

At #x=3#, the derivative is:

#dy/dx|_(x=3)# #=1/(2sqrt(3+1))=1/4#

So, the tangent line has slope #1//4# and passes through the point #(3,2)#. In point-slope form:

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

#y-2=1/4(x-3)#

Or:

#y=1/4x+5/4#

The graph of the function and its tangent line:

graph{(y-sqrt(x+1))(y-(1/4x+5/4))=0 [-1.11, 14.69, -2.39, 5.51]}