y'?
1.y=arctan(a+x/1-ax); a is constant
2.y=e^x-e^-x/e^x+e^-x
recall that if
y = arctan(u)
y' = 1/(1+u^2) u'
y=arctan(a+x/1-ax)
y = 1/(1+((a+x)/(1-ax))^2) * (1+a^2)/(1-ax)^2
= ...
amazing!
y=(e^x-e^-x)/(e^x+e^-x) = tanh(x)
...
#1 is not quite so amazing if you recognize that if
a = tan(u)
x = tan(v)
y = arctan(tan(u+v)) = u+v
y' = v'
...
To find the derivative of the given functions, we can use basic differentiation rules. Here's how you can find the derivative of each function:
1. y = arctan(a + x / 1 - ax)
To find the derivative of y, we can use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) is f'(g(x)) * g'(x).
Let's break down the function:
- f(u) = arctan(u), where u = a + x / (1 - ax)
- g(x) = a + x / (1 - ax)
Now, we can find the derivatives of f(u) and g(x) separately:
- f'(u) = 1 / (1 + u^2) [derivative of arctan(u)]
- g'(x) = 1 - a^2 / (1 - ax)^2 [derivative of (a + x) / (1 - ax)]
By the chain rule, the derivative of y is:
y' = f'(g(x)) * g'(x) = [1 / (1 + (a + x / (1 - ax))^2)] * [1 - a^2 / (1 - ax)^2]
Simplifying further may depend on the level of detail or form you want for the answer.
2. y = (e^x - e^(-x)) / (e^x + e^(-x))
To find the derivative of y, let's rewrite the expression as:
y = (e^x - e^(-x)) * (e^x + e^(-x))^(-1)
Now, we can differentiate using the quotient rule, which states that if we have a quotient of two functions, f(x) and g(x), the derivative is (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2.
Let's break down the function:
- f(x) = e^x - e^(-x)
- g(x) = e^x + e^(-x)
Now, we can find the derivatives of f(x) and g(x) separately:
- f'(x) = e^x + e^(-x) [derivative of e^x - e^(-x)]
- g'(x) = e^x - e^(-x) [derivative of e^x + e^(-x)]
Applying the quotient rule, the derivative of y is:
y' = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2
= [(e^x + e^(-x)) * (e^x + e^(-x)) - (e^x - e^(-x)) * (e^x - e^(-x))] / (e^x + e^(-x))^2
In this case, we don't have any further simplification, unless specified.