Find the derivative dy/dx
y=e^(sinh(x))
just use the good old chain rule. By now that shouldn't be causing any difficulties, eh?
If y = e^u, y' = e^u u'
y = e^(sinh(x))
y' = e^sinh(x)) * cosh(x)
To find the derivative of y = e^(sinh(x)), we can use the chain rule.
Let's denote u = sinh(x). Then, y = e^u.
Now, we can find the derivative using the chain rule:
dy/dx = (dy/du) * (du/dx)
To find dy/du:
Since y = e^u, the derivative of y with respect to u is simply e^u.
dy/du = e^u
Next, let's find du/dx:
Since u = sinh(x), we can rewrite it as u = (e^x - e^(-x)) / 2.
Let's take the derivative of u with respect to x:
du/dx = (e^x + e^(-x)) / 2
Now, we can substitute dy/du and du/dx into the chain rule equation:
dy/dx = (dy/du) * (du/dx)
= e^u * ((e^x + e^(-x)) / 2)
Since u = sinh(x), we can substitute it back in:
dy/dx = e^(sinh(x)) * ((e^x + e^(-x)) / 2)
= ((e^x + e^(-x)) / 2) * e^(sinh(x))
Therefore, the derivative of y = e^(sinh(x)) is dy/dx = ((e^x + e^(-x)) / 2) * e^(sinh(x)).
To find the derivative of y = e^(sinh(x)), we can use the chain rule. The chain rule states that if we have a function y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).
Let's break down the function y = e^(sinh(x)) using the chain rule. In this case, f(u) = e^u and g(x) = sinh(x).
First, we need to find the derivative of the outer function f(u) = e^u. The derivative of e^u is itself, so f'(u) = e^u.
Next, we need to find the derivative of the inner function g(x) = sinh(x). The derivative of sinh(x) is cosh(x), so g'(x) = cosh(x).
Now we can apply the chain rule by multiplying the derivatives:
dy/dx = f'(g(x)) * g'(x)
= e^(sinh(x)) * cosh(x)
Therefore, the derivative of y = e^(sinh(x)) is dy/dx = e^(sinh(x)) * cosh(x).