Question

Find the slope of the line tangent to the curve 2^x + 3x^2y+y = 1 at the point (0,1).
Can anyone walk me through this?

Answer 1

\[2^x+3x^2y+y = 1\]
The slope of the line tangent to the curve of this?

Answer 2

The line tangent to the curve is represented by the derivative function evaluated at the given point.

So you'll need to take the derivative `implicitly`.

Answer 3

So you have to apply a few different rules to take the derivative of each term.
Remember how to deal with an exponential?\[\Large\rm \frac{d}{dx}2^x=?\]

Answer 4

Not really, no.

Answer 5

Mmm ok well maybe go review that at some point.
You get the same thing back, but multiplied by the log of the base.

Maybe you'll recall this derivative:\[\Large\rm \frac{d}{dx}e^x=e^x\]It's more accurately this (following the exponential rule for derivatives):\[\Large\rm \frac{d}{dx}e^x=e^x(\ln e)\]But since lne=1, we simply ignore it.

Answer 6

For the second term, you need to apply the product rule,
\[\Large\rm \left(3x^2y\right)'=\color{royalblue}{\left(3x^2\right)'}y+3x^2\color{royalblue}{\left(y\right)'}\]