Find the derivative of the given function.
[tex]f(x)=\cos x \cot x[/tex]
[tex]f^{\prime}(x)=-[?] x \quad \square^2 x+\square x \cot x[/tex]
Answer (2)
Apply the product rule to f ( x ) = cos x cot x , where u = cos x and v = cot x .
Find the derivatives u ′ = − sin x and v ′ = − csc 2 x .
Substitute into the product rule: f ′ ( x ) = ( − sin x ) ( cot x ) + ( cos x ) ( − csc 2 x ) .
Simplify to get f ′ ( x ) = − cos x csc 2 x − cos x , so the missing terms are cos x and cos x .
Explanation
Problem Analysis We are given the function f ( x ) = cos x cot x and asked to find its derivative f ′ ( x ) . We need to express the derivative in the form f ′ ( x ) = − [ ?] cos x csc 2 x + [ ?] cos x .
Applying the Product Rule To find the derivative, we will use the product rule. Let u ( x ) = cos x and v ( x ) = cot x . Then f ( x ) = u ( x ) v ( x ) , and f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) .
Finding Derivatives We find the derivatives of u ( x ) and v ( x ) .
u ′ ( x ) = d x d ( cos x ) = − sin x
v ′ ( x ) = d x d ( cot x ) = − csc 2 x
Substituting into the Product Rule Now, we substitute these into the product rule formula:
f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) = ( − sin x ) ( cot x ) + ( cos x ) ( − csc 2 x )
Simplifying the Expression Next, we simplify the expression. Recall that cot x = s i n x c o s x and csc x = s i n x 1 .
f ′ ( x ) = − sin x ⋅ s i n x c o s x − cos x ⋅ s i n 2 x 1 = − cos x − s i n 2 x c o s x
Rewriting the Expression We can rewrite the second term using csc 2 x = s i n 2 x 1 :
f ′ ( x ) = − cos x − cos x csc 2 x
Factoring Finally, we factor out − cos x :
f ′ ( x ) = − cos x ( 1 + csc 2 x ) = − cos x csc 2 x − cos x
Final Answer Comparing this with the desired form f ′ ( x ) = − [ ?] cos x csc 2 x + [ ?] cos x , we see that the missing terms are 1 and − 1 .
Thus, f ′ ( x ) = − cos x csc 2 x − cos x .
Conclusion Therefore, the derivative of f ( x ) = cos x cot x is f ′ ( x ) = − cos ( x ) csc 2 ( x ) − cos ( x ) .
Examples
In physics, understanding derivatives of trigonometric functions is crucial for analyzing oscillatory motion, such as the motion of a pendulum or the vibration of a string. The derivative helps determine the velocity and acceleration of the oscillating object, providing insights into its dynamic behavior. For example, if you're designing a suspension system for a car, knowing how the displacement changes with time (velocity) and how the velocity changes with time (acceleration) is essential for optimizing the system's performance and ensuring a smooth ride.
The derivative of the function f ( x ) = cos x cot x is found using the product rule. The final result is f ′ ( x ) = − cos x csc 2 x − cos x . The missing coefficients are 1 and − 1 .
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