Question

find the equation of the tangent line to the graph of f(x)=ln(1+x^3) at the point (0,0)

Answer 1

What's the derivative of f(x)?

Answer 2

1 right ?

Answer 3

\[\large \frac{d}{dx}f(x)=\frac{d}{dx}\log(1+x^3)\]

Answer 4

How do you make fractions? (not the kind with the slanted bar)

Answer 5

\[\large =\frac{d}{dx}f(x)\\\large=\frac{d}{dx}\ln(1+x^3)\\\large= \frac{1}{1+x^3}*f'(1+x^3)\]

Answer 6

Use power rule
\[\large= \frac{1}{1+x^3}*f'(1+x^3)\\\large =\frac{3x^2}{1+x^3}\]

Answer 7

Oh, is it just \frac{a}{b}?

Answer 8

\[\large f'(x)=\frac{3x^2}{1+x^3}\]A tangent line is a linear equation, \(\large y=mx+b\), you want the equation of the line at point (0,0). Plug in 0 in to find the slope;
\[\large\frac{3x^2}{1+x^3}=\frac{3(0)^2}{1+(0)^3}=0\]

Answer 9

Now that's easy, just find the equation of the tangent line:
\[\large y=mx+b\\\large 0=0x+b\\\large0=0+b\\\large b=0\]

Answer 10

Therefore equation of the tangent line is \(\large y=0\)

Answer 11

ok I got it.

Answer 12

Sorry, should be \[\large \begin{align}
&=\frac{1}{1+x^3}*\frac{d}{dx}(1+x^3)\\
&=\frac{3x^2}{1+x^3}
\end{align}\]here: http://puu.sh/K2g3

A little mistake in how I wrote the derivative thing.