Question

Find the equation of the tangent line to the graph of f(x)= ln(1+x^2) at the point (0,0)

Answer 1

Can you find the derivative of this function?

Answer 2

Find derivative and then put it in the eqn:
(y-y1) = m(x-x1)

Answer 3

where m = derivative
x = 0
y = 0

Answer 4

\[f'(0)=m \\ (x_a,y_b)=(0,0) \\ y-y_a=m(x-x_a)\]

Answer 5

The only real work that it tough is finding the derivative of
\[f(x)= \ln(1+x^2) \]Think you can find it?

Answer 6

\[\large \frac{dy}{dx}f(x)=\ln(1+x^2)=\frac{\frac{dy}{dx}(1+x^2)}{1+x^2}=\frac{2x}{1+x^2}\]

Answer 7

ok the derivative of this function is (derivatives gives us slope)
\[m=f'(x)=\frac{2x}{1+x^2}\]
put x= 0 in the above to get numerical value of slope
\[m=\frac{2*0}{1+0^2}=0\]

so slope is zero!
and you may know that the slope of horizontal line is zero whose equation is
!

Answer 8

did u know the equation of horizontal line??

Answer 9

and the asker is just looking around :(

Answer 10

@cunninnc

Answer 11

@sami-21 yes thank you

Answer 12

to find the tangent line of this eq. \[f(x)=e ^{x}/2x+1 \] would my derivative be \[f \prime= (2x+1)(e ^{x}) - (e ^{x}) (2) / (2x+1)^{2} \]

Answer 13

\[f′=\frac{(2x+1)(e^x)−(e^x)(2)}{(2x+1)^2}\]

Answer 14

Yup you are right

Answer 15

thanks so do i leave it like that or simplify

Answer 16

Do i simply solve by pluggin in?