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Derivative of Trigonometric Functions Video Lecture | Mathematics for Grade 11

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FAQs on Derivative of Trigonometric Functions Video Lecture - Mathematics for Grade 11

1. What is the derivative of the sine function?
Ans. The derivative of the sine function, denoted as d/dx(sin(x)), is equal to the cosine function, or cos(x). In other words, the rate of change of the sine function at any given point is given by the cosine function.
2. What is the derivative of the cosine function?
Ans. The derivative of the cosine function, denoted as d/dx(cos(x)), is equal to the negative sine function, or -sin(x). This means that the rate of change of the cosine function at any given point is given by the negative sine function.
3. How do you differentiate the tangent function?
Ans. To differentiate the tangent function, we use the quotient rule. The derivative of the tangent function, denoted as d/dx(tan(x)), is equal to sec^2(x), where sec(x) represents the secant function. Therefore, the rate of change of the tangent function at any given point is given by the secant squared function.
4. What is the derivative of the cotangent function?
Ans. The derivative of the cotangent function, denoted as d/dx(cot(x)), can be obtained by differentiating the reciprocal of the tangent function. Using the quotient rule, we find that the derivative of the cotangent function is equal to -csc^2(x), where csc(x) represents the cosecant function. Hence, the rate of change of the cotangent function at any given point is given by the negative cosecant squared function.
5. How do you find the derivative of the secant function?
Ans. To find the derivative of the secant function, denoted as d/dx(sec(x)), we can use the product rule in combination with the chain rule. The derivative is given by d/dx(sec(x)) = sec(x)tan(x). This means that the rate of change of the secant function at any given point is equal to the secant function multiplied by the tangent function.
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