1)d/dx of cos4x/3-x^2(inculding quentient&chain rule)
2)sin^2x/1+cosx(same method)
#1
y = cos 4x/(3-x^2)
dy/dx = ( (3-x^2)(-4sin 4x) - cos 4x (-2x) )/(3-x^2)^2
try the second one, let me know what you got
you might be surprised at the answer to #2, unless you note that
sin^2/(1+cos) = (1-cos^2)/(1+cos) = 1-cos
no need for quotient rule or chain rule here.
To find the derivative of these functions, we will need to use the quotient rule and the chain rule.
1) d/dx of (cos(4x))/(3 - x^2):
The quotient rule states that if we have a function u(x) divided by another function v(x), then the derivative can be found using the formula:
d/dx (u(x)/v(x)) = (v(x) * u'(x) - u(x) * v'(x))/(v(x))^2
In this case, u(x) = cos(4x) and v(x) = 3 - x^2.
First, we need to find the derivatives of u(x) and v(x) individually, which involves using the chain rule.
The chain rule states that if we have a composition of functions, such as f(g(x)), then the derivative can be found using the formula:
d/dx (f(g(x))) = f'(g(x)) * g'(x)
Let's proceed step by step:
u(x) = cos(4x) => u'(x) = -4sin(4x) (Since the derivative of cos(x) is -sin(x), and we have a chain rule with g(x) = 4x)
v(x) = 3 - x^2 => v'(x) = -2x (Since the derivative of x^2 is 2x)
Now, let's substitute the values into the quotient rule formula:
d/dx (cos(4x))/(3 - x^2) = ((3 - x^2) * (-4sin(4x)) - (cos(4x)) * (-2x))/((3 - x^2))^2
Simplifying further, we have:
d/dx (cos(4x))/(3 - x^2) = (-4(3 - x^2)sin(4x) + 2x(cos(4x)))/((3 - x^2))^2
Therefore, the derivative of (cos(4x))/(3 - x^2) is (-4(3 - x^2)sin(4x) + 2x(cos(4x)))/((3 - x^2))^2.
2) d/dx of (sin^2(x))/(1 + cos(x)):
Using the quotient rule, we have:
d/dx (sin^2(x))/(1 + cos(x)) = ((1 + cos(x)) * (2sin(x) * cos(x)) - (sin^2(x)) * (-sin(x)))/((1 + cos(x))^2)
Simplifying further, we have:
d/dx (sin^2(x))/(1 + cos(x)) = (2sin(x) * cos^2(x) + sin^3(x))/((1 + cos(x))^2)
Therefore, the derivative of (sin^2(x))/(1 + cos(x)) is (2sin(x) * cos^2(x) + sin^3(x))/((1 + cos(x))^2).