Science:Math Exam Resources/Courses/MATH102/December 2019/Question 05 (a)/Solution 1

The slope of the tangent line to the curve $y=f(x)$ is equal to $f'(x)$.

$f'(10)={\frac {1}{(1+10)^{2}}}={\frac {1}{121}}$. Using the slope-point form, the tangent line equation is

$y-{\frac {10}{11}}={\frac {1}{121}}\left(x-10\right)$,

i.e.

$y={\frac {1}{121}}x+{\frac {100}{121}}$.