Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx(-2 sin(sqrt(tan(sin(x)))))
The derivative of y is y'(x):
y'(x) = d/dx(-2 sin(sqrt(tan(sin(x)))))
Factor out constants:
y'(x) = -2 d/dx(sin(sqrt(tan(sin(x)))))
Using the chain rule, d/dx(sin(sqrt(tan(sin(x))))) = ( dsin(u))/( du) 0, where u = sqrt(tan(sin(x))) and ( d)/( du)(sin(u)) = cos(u):
y'(x) = -2 cos(sqrt(tan(sin(x)))) d/dx(sqrt(tan(sin(x))))
Using the chain rule, d/dx(sqrt(tan(sin(x)))) = ( dsqrt(u))/( du) 0, where u = tan(sin(x)) and ( d)/( du)(sqrt(u)) = 1/(2 sqrt(u)):
y'(x) = -2 cos(sqrt(tan(sin(x)))) (d/dx(tan(sin(x))))/(2 sqrt(tan(sin(x))))
Simplify the expression:
y'(x) = -(cos(sqrt(tan(sin(x)))) (d/dx(tan(sin(x)))))/sqrt(tan(sin(x)))
Using the chain rule, d/dx(tan(sin(x))) = ( dtan(u))/( du) 0, where u = sin(x) and ( d)/( du)(tan(u)) = sec^2(u):
y'(x) = -(cos(sqrt(tan(sin(x)))))/sqrt(tan(sin(x))) d/dx(sin(x)) sec(sin(x))^2
The derivative of sin(x) is cos(x):
y'(x) = -(cos(sqrt(tan(sin(x)))) sec^2(sin(x)))/sqrt(tan(sin(x))) cos(x)
Expand the left hand side:
Answer: |y'(x) = -(cos(x) cos(sqrt(tan(sin(x)))) sec^2(sin(x)))/sqrt(tan(sin(x)))
