Write the composite function in the form f(g(x)). [Identify the inner function u=g(x) and the outer function y=f(u).] Then find the derivative dy/dx.
y=√sinx
√ is square root.
g(x) = sinx
f(x) = √x
then f(g(x))
= f(sinx)
= √sinx
if y = fg(x)) = √sinx = (sinx)^(1/2)
dy/dx = (1/2)(sinx)^(-1/2) (cosx
= cosx/(2√sinx)
To write the composite function in the form f(g(x)), let's first identify the inner function u as g(x) and the outer function y as f(u).
Given: y = √sinx
Inner function: u = g(x) = sinx
Outer function: y = f(u) = √u
Now we can express the given function as y = f(g(x)):
y = √sinx = f(g(x))
To find the derivative dy/dx, we can use the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then the derivative dy/dx is given by dy/dx = f'(g(x)) * g'(x).
Let's find the derivative of y = √u with respect to u first:
dy/du = d(√u)/du
To differentiate the square root function, we use the power rule. If we have a function y = √u, then dy/du = (1/2)u^(-1/2):
dy/du = (1/2)u^(-1/2)
Now, let's find the derivative of the inner function u = sinx with respect to x:
du/dx = d(sinx)/dx = cosx
Using the chain rule, we can find dy/dx:
dy/dx = (dy/du) * (du/dx)
= (1/2)u^(-1/2) * cosx
Substituting the values of u and simplifying, we get:
dy/dx = (1/2)(sinx)^(-1/2) * cosx
So, the derivative of y = √sinx with respect to x is dy/dx = (1/2)(sinx)^(-1/2) * cosx.
To write the composite function in the form f(g(x)), let's first identify the inner function u=g(x) and the outer function y=f(u).
In this case, the inner function is u = g(x) = x, and the outer function is y = f(u) = √u.
Now, let's find the derivative dy/dx using the chain rule.
1. First, find the derivative of the outer function y = √u with respect to its variable u:
dy/du = 1/(2√u)
2. Next, find the derivative of the inner function u = x with respect to its variable x:
du/dx = 1
3. Now, applying the chain rule, we can find the derivative of the composite function y = √sinx with respect to x:
dy/dx = (dy/du) * (du/dx)
Substituting the derivatives we found:
dy/dx = (dy/du) * (du/dx)
= (1/(2√u)) * 1
= 1/(2√u)
Since our inner function is u = g(x) = x, we can substitute it back:
dy/dx = 1/(2√x)
Therefore, the derivative of y = √sinx with respect to x is dy/dx = 1/(2√x).