How do you find the equation of the tangent line to the graph of #f(x)=sqrtx# at point (1,1)?

Question

Aurora Alvarado · Accepted Answer

#y=1/2x+1/2# The equation of the tangent in #color(blue)"point-slope form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))# where m represents the slope and # (x_1,y_1)" is a point on the tangent"# #color(orange)"Reminder: " m_("tangent")=dy/dx" at x=a"# #f(x)=sqrtx=x^(1/2)# differentiate using the #color(blue)"power rule"# #rArrf'(x)=1/2x^(-1/2)=1/(2sqrtx)# #rArrf'(1)=1/2=m_("tangent")# #"Using " m=1/2" and " (x_1,y_1)=(1,1)" then"# #rArry-1=1/2(x-1)# #rArry=1/2x-1/2+1# #rArry=1/2x+1/2" is equation of tangent line"#

Samantha Adkison · Answer

undefined To find the equation of the tangent line to the graph of f(x) = √x at point (1,1), we can use the concept of differentiation.

First, we need to find the derivative of f(x). The derivative of √x is 1/(2√x).

Next, we substitute x = 1 into the derivative to find the slope of the tangent line at point (1,1). 
So, the slope of the tangent line is 1/(2√1) = 1/2.

Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values, we get:
y - 1 = (1/2)(x - 1)

Simplifying the equation, we have:
y - 1 = (1/2)x - 1/2

Rearranging the equation, we get the equation of the tangent line:
y = (1/2)x + 1/2