Differentiate f(x)=arctan(e^5x)

In summary, the conversation was about differentiating the function f(x)=arctan(e^5x) using the chain rule. The correct derivative is 5e^5x/(1+(e^10x)). The conversation also mentions helpful websites for finding derivatives and integrals.
  • #1
jpc90
8
0
How do you differentiate f(x)=arctan(e^5x)
 
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  • #2
If it's w.r.t . x then use the chain rule e.g. u= e^(5x)
 
  • #3
I did something like,

1/1+(e^5x)^2

and then

5e^-5x/1+(e^10x)

I don't think that is right though, I'm pretty confused
 
Last edited:
  • #4
How did you get e-5x in the numerator? The derivative of e5x is 5e5x. That is what you should be multiplying by.
 
  • #5
I thought since e^5x is in the denominator that it is equal to e^-5x
 
  • #6
Last edited by a moderator:
  • #7
ok thanks guys
 

1. What is the derivative of f(x) = arctan(e^5x)?

The derivative of f(x) = arctan(e^5x) is given by f'(x) = (e^5x)/(1 + e^(10x)). This can be derived using the chain rule and the derivative of arctan function.

2. How do you find the critical points of f(x) = arctan(e^5x)?

To find the critical points of f(x) = arctan(e^5x), we need to find the values of x where f'(x) = 0. This will give us the stationary points of the function, which can be considered as critical points.

3. Is f(x) = arctan(e^5x) a continuous function?

Yes, f(x) = arctan(e^5x) is a continuous function. Both the arctan function and the exponential function are continuous, and the composition of continuous functions is also continuous.

4. What is the range of f(x) = arctan(e^5x)?

The range of f(x) = arctan(e^5x) is (-π/2, π/2). This is because the range of the arctan function is (-π/2, π/2), and the range of e^5x is (0, ∞). The composition of these two functions will give us the intersection of their ranges.

5. How do you sketch the graph of f(x) = arctan(e^5x)?

To sketch the graph of f(x) = arctan(e^5x), we can use the properties of the arctan function and the exponential function. We can also plot a few points and use them to draw a smooth curve. The graph will have a horizontal asymptote at y = π/2 and a vertical asymptote at x = 0.

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