Find the derivative of y=cos(squarerootsin(tzn(pi*x)))
To find the derivative of the function y = cos(sqrt(sin(t*z*n(pi*x)))), we need to apply the chain rule. The chain rule states that if we have a composite function like f(g(x)), then its derivative can be computed by taking the derivative of the outer function (f') and multiplying it by the derivative of the inner function (g').
Let's break down the function and calculate the derivative step by step:
1. Identify the inner function: g(x) = sqrt(sin(t*z*n(pi*x))).
2. Compute the derivative of the inner function g'(x):
To differentiate sqrt(sin(t*z*n(pi*x))), we need to apply the chain rule again since we have the square root of the sine function.
Let's break it down further:
Let u = sin(t*z*n(pi*x)).
Then, our function becomes g(x) = sqrt(u).
To find g'(x), we apply the chain rule:
g'(x) = (1/2)*u^(-1/2)*u',
where u' is the derivative of u with respect to x.
To find u', we apply the chain rule again:
Let v = t*z*n(pi*x).
u' = cos(v)*v',
where v' is the derivative of v with respect to x.
Applying the chain rule again, we have:
v' = t*z*n(pi).
Therefore, u' = cos(t*z*n(pi*x))*v' = cos(t*z*n(pi*x))*t*z*n(pi).
Now, substituting u' back into g'(x):
g'(x) = (1/2)*u^(-1/2)*cos(t*z*n(pi*x))*t*z*n(pi).
3. Identify the outer function: f(x) = cos(x).
4. Compute the derivative of the outer function f'(x):
f'(x) = -sin(x).
5. Apply the chain rule:
Now that we have g'(x) and f'(x), we can use the chain rule to find the derivative of the original function (y = cos(sqrt(sin(t*z*n(pi*x))))).
dy/dx = f'(g(x)) * g'(x).
Substituting f'(x) = -sin(x) and g'(x) = (1/2)*u^(-1/2)*cos(t*z*n(pi*x))*t*z*n(pi), we get:
dy/dx = -sin(sqrt(sin(t*z*n(pi*x)))) * (1/2)*[sin(t*z*n(pi*x))]^(-1/2) * cos(t*z*n(pi*x)) * t * z * n(pi).
Therefore, the derivative of y = cos(sqrt(sin(t*z*n(pi*x)))) is dy/dx = -sin(sqrt(sin(t*z*n(pi*x)))) * (1/2)*[sin(t*z*n(pi*x))]^(-1/2) * cos(t*z*n(pi*x)) * t * z * n(pi).