Differentiation of an exponential function

AI Thread Summary
To find the coordinates of the point on y = e^x where the gradient is 2, the derivative of e^x, which is e^x itself, must be set equal to 2. This leads to the equation e^x = 2, which can be solved using natural logarithms. The solution involves taking the natural log of both sides to find x = ln(2). After determining x, the corresponding y-coordinate can be calculated by substituting back into the original function. Understanding that the derivative of e^x is simply e^x is crucial for solving this problem.
pavadrin
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Hey,
I have a problem involving natural logs which has got me confused, even though it appears simple.
The problem: Find the exact coordinates of the point on y = e^x where the gradient is 2.
From previous experience, I know that differentiation is required, but because of the e I am not sure on how to go about this. After the differentiation I think I can manage to complete it.
Thanks,
Pavdarin
 
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pavadrin said:
Hey,
I have a problem involving natural logs which has got me confused, even though it appears simple.
The problem: Find the exact coordinates of the point on y = e^x where the gradient is 2.
From previous experience, I know that differentiation is required, but because of the e I am not sure on how to go about this. After the differentiation I think I can manage to complete it.
Thanks,
Pavdarin

Well what is the derivative of y = e^x?
 
if you don't know it you can find out using the given that d/dx ln x = 1/x and then calculate d/dx ln e^x using the chain rule
 
If you know how to do this kind of problem then all you need is the derivative of e^x which is just e^x. Just do this problem as you would do if "y" was any other function, for example a polynomial.
 
thanks for the help, however i am still unsure on how to approach this problem
 
pavadrin said:
thanks for the help, however i am still unsure on how to approach this problem

the derivative is the gradient, so the following must be solved:

\frac {dy} {dx}\ = 2

This will give you the x-coordinate.
 
Last edited:
pavadrin said:
thanks for the help, however i am still unsure on how to approach this problem

You "approach" this problem by differentiating ex! What is the derivative of ex? (It's the world's easiest derivative!)
 
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