Question

find an equation of the tangent line to
the graph of the function at the given point.

Answer 1

We can write the equation of a line in Slope-Intercept form, \(\large y=mx+b\)
Or in Point-Slope form, \(\large y-y_o=m(x-x_o)\)

Which form are you more familiar working with?

Answer 2

y=mx+b

Answer 3

Ok cool.
Then we have two pieces of information we need to find in order to establish the equation of this line.

We'll need the slope of the line, \(\large m\)
and the y-intercept, \(\large b\)

Answer 4

The slope of the line is given by the derivative of the function at the given point.

\(\large m=f'(1)\)

Answer 5

To find the derivative, use rules of logarithms first.
We don't want to have to apply the quotient rule. it will get really messy.

Answer 6

\[\large \ln\left(\frac{a}{b}\right) \qquad = \qquad \ln(a)-\ln(b)\]

Answer 7

Applying this gives us,
\[\large y=\ln\left(\frac{1+x}{x}\right) \qquad \rightarrow \qquad y=\ln(1+x)-\ln(x)\]
Will be much easier to take a derivative from here :)
Can you take the derivative of this?

Answer 8

yes hold on

Answer 9

dy/dn=1

Answer 10

? D:

Answer 11

\[\large y'=\frac{1}{1+x}-\frac{1}{x}\]

\[\large y'(1)=\frac{1}{1+1}-\frac{1}{1}\]

Answer 12

i know lol thats what i got @zepdrix