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

1 Answer
Jan 31, 2017

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

Explanation:

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"#