The slope of the tangent line to the curve y = f ( x ) {\displaystyle y=f(x)} is equal to f ′ ( x ) {\displaystyle f'(x)} .
f ′ ( 10 ) = 1 ( 1 + 10 ) 2 = 1 121 {\displaystyle f'(10)={\frac {1}{(1+10)^{2}}}={\frac {1}{121}}} . Using the slope-point form, the tangent line equation is
y − 10 11 = 1 121 ( x − 10 ) {\displaystyle y-{\frac {10}{11}}={\frac {1}{121}}\left(x-10\right)} ,
i.e.
y = 1 121 x + 100 121 {\displaystyle y={\frac {1}{121}}x+{\frac {100}{121}}} .
is equal to 
.
. Using the slope-point form, the tangent line equation is
,
.
