Now use the product rule
Find an equation for the line tangent to the graph of f(x)=3x(e^x) at the point (a, f(a)) if a=2.
I think you start with taking a derivative. Isn't the derivative just f'(x)=3e^x or do you have to do the product rule? After that, I'm not sure what to do.
Sorry to keep asking questions, but I worked it out and still didn't get it right.
So I plugged 2 into the derivative and got 66.50150489, and then I plugged in a and f(a) and found C to be .3334012213, so I thought my equation would be
y=66.50150489(x+.3334012213), but the math program I'm using is telling me that this is incorrect.
If a = 2 then and the slope of the tangent is:
You know a point and the slope of the tangent. Use the point-slope-formula to calculate the equation of the straight line:
As you may have noticed: The terms I use are a lot shorter than the decimal numbers you use - and they are exact! The decimal numbers are only approximative values.