- #1
Emjay
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Could I please get help with the following question?
f(x)=(2cos^2 x+3)^5/2
Any help would be very much appreciated:)
f(x)=(2cos^2 x+3)^5/2
Any help would be very much appreciated:)
Emjay said:Yes, that is correct :)
Honestly, not even sure where to start with this one.
I have only ever done basic examples and this one has got me stumped.
The basic trigonometric functions are sine, cosine, and tangent. They are commonly denoted as sin(x), cos(x), and tan(x) respectively.
To find the derivative of a trigonometric function, you can use the following rules:
- The derivative of sine is cosine (d/dx(sin(x)) = cos(x))
- The derivative of cosine is negative sine (d/dx(cos(x)) = -sin(x))
- The derivative of tangent is secant squared (d/dx(tan(x)) = sec^2(x))
Yes, you can use the chain rule to find the derivative of a trigonometric equation. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). This can be applied to trigonometric equations by substituting in the appropriate derivative for the inner function.
Yes, there are a few special cases when finding the derivative of a trigonometric function. For example, the derivative of secant is secant*tangent (d/dx(sec(x)) = sec(x)*tan(x)). Also, the derivative of cotangent is negative cosecant squared (d/dx(cot(x)) = -csc^2(x)).
To find the derivative of a trigonometric equation with multiple trigonometric functions, you can use the product rule or quotient rule depending on the form of the equation. The product rule states that if y = f(x)*g(x), then dy/dx = f'(x)*g(x) + f(x)*g'(x). The quotient rule states that if y = f(x)/g(x), then dy/dx = (f'(x)*g(x) - f(x)*g'(x))/(g(x))^2.