How do you find the equation of a line tangent

Salma Meza

Salma Meza

Answered question

2022-02-10

How do you find the equation of a line tangent to the function y=ln(x) at (2,ln2)?

Answer & Explanation

liiipstick0j2

liiipstick0j2

Beginner2022-02-11Added 14 answers

Let's first determine whether (2,ln2) lies on y=ln(x), Put x=-2 in y=ln(x), we get ln2 and hence it lies on the currve. 
Now, the slope of a curve determined by y=f(x) is determined by  dy  dx  
As y=ln(x) dy  dx =1x×1=1x 
and at x=-2,  dy  dx =12 
As tangent has a slope 12 and passes through (2,ln2), its equation is 
yln2=12(x(2)) 
or 2y2ln2=x2 
or x+2y+22ln2=0

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