Question

If \(x=\tan t+\cot t, y=2 \log (\cot t)\) then \((d y / d x)=\) (a) \(-\tan 2 t\) (b) \(\tan 2 \mathrm{t}\) (c) \(\sin 2 \mathrm{t}\) (d) \(\cos 2 t\)

Short answer

The correct answer is \((a)\) \(-\tan 2t\).

Step-by-step solution

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

We are given: \[x = \tan t + \cot t\] \[y = 2\log(\cot t)\] First, let's find \(\frac{dx}{dt}\). To do this, we will find the derivatives of both \(\tan t\) and \(\cot t\). \[ \frac{d}{dt} (\tan t) = \sec^2 t \] \[ \frac{d}{dt} (\cot t) = -\csc^2 t\] Using these, \(\frac{dx}{dt}\) is: \[ \frac{dx}{dt} = \sec^2 t - \csc^2 t\] Next, let's find \(\frac{dy}{dt}\). To do this, we need to find the derivative of \(2\log(\cot t)\). \[ \frac{d}{dt} (2\log(\cot t)) = 2\frac{1}{\cot t}\cdot (-\csc^2 t) \] \[ \frac{dy}{dt} = -2\csc^2 t\tan t \] Now we have the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

Apply the chain rule to find \(\frac{dy}{dx}\):

To find \(\frac{dy}{dx}\), we can use the chain rule, that is: \[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \] Notice that from our previous step, to get \(\frac{dt}{dx}\), we simply need to find the reciprocal of \(\frac{dx}{dt}\): \[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{\sec^2 t - \csc^2 t} \] Now, we can find \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = (-2\csc^2 t\tan t) \cdot \frac{1}{\sec^2 t - \csc^2 t} \] Simplify the expression: \[ \frac{dy}{dx} = -\frac{2\csc^2 t\tan t}{\sec^2 t - \csc^2 t} \] Now, let's analyze the expression to check if it matches any of the given options.

Match the solution with the given options and determine the answer:

Let's rewrite our expression using sin and cos functions, and simplify further: \[ \frac{dy}{dx} = -\frac{2\cdot \frac{1}{\sin^2 t}\cdot \frac{\sin t}{\cos t}}{\frac{1}{\cos^2 t}-\frac{1}{\sin^2 t}}\] \[ \frac{dy}{dx} = -\frac{2\cdot \frac{1}{\sin t}\cdot \frac{\sin t}{\cos t}}{\frac{\sin^2 t - \cos^2 t}{\sin^2 t\cos^2 t}}\] \[ \frac{dy}{dx} = -\frac{2(\sin^2 t\cos^2 t)}{(\sin t \cos t)(\sin^2 t -\cos^2 t)} \] \[ \frac{dy}{dx} = -\frac{2\cos t \sin t}{\sin^2 t - \cos^2 t} \] Observe that this expression matches option (a) if we use the double angle identity for sine: \(\sin 2t = 2\sin t \cos t\). Therefore: \[ \frac{dy}{dx} = -\tan 2t\] The correct answer is (a) \(-\tan 2t\).

Key concepts

The Chain Rule

One of the most essential concepts in differential calculus is the chain rule. It is crucial when we need to find the derivative of a composite function. A composite function can be thought of as a function within another function, and understanding how to differentiate these comfortably is a key skill in calculus.

The chain rule states that if you have two functions, say, f(g(x)), the derivative of this composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In mathematical terms, this is expressed as \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).

When confronted with a situation where you need to find \( \frac{dy}{dx} \) but you are given functions in terms of another variable t, as in our original problem, the chain rule helps us to form a bridge. We find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) and then multiply \( \frac{dy}{dt} \) by the reciprocal of \( \frac{dx}{dt} \) which effectively gives us \( \frac{dy}{dx} \), just as shown in the textbook solution step.

Trigonometric Derivatives

Trigonometry plays a vital role in calculus, especially when working with periodic functions. Derivatives of trigonometric functions are common in problems, and understanding them is crucial for calculating more complex derivatives and integrals.

Some basic trigonometric derivatives that any calculus student must know include:

• The derivative of \( \sin x \) is \( \cos x \)
• The derivative of \( \cos x \) is \( -\sin x \)
• The derivative of \( \tan x \) is \( \sec^2 x \)
• The derivative of \( \cot x \) is \( -\csc^2 x \)

These were involved in the solution process of our given problem. Recognizing how to take these derivatives quickly and accurately is key to efficiency in calculus problems, especially when combined with the chain rule for more complex functions.

Also, knowing the trigonometric identities, such as the double angle formula which was used in the final step to match the solution to given options, can dramatically simplify the process of finding the desired derivatives.

Implicit Differentiation

When functions are given in an explicit form, say y as a function of x, differentiation is straightforward. But often, we encounter equations where y is not explicitly isolated and is tangled with x in some way - this is where implicit differentiation comes into play.

In implicit differentiation, rather than solving the equation for y to then differentiate, we differentiate both sides of the equation with respect to x, keeping in mind that y is a function of x. This means that whenever we differentiate y, we multiply the result by \( \frac{dy}{dx} \) because of the chain rule.

In the exercise we discussed, implicit differentiation wasn't used directly, as we had separate functions for x and y in terms of t. However, knowing this technique is invaluable when dealing with equations where separating the variables is not easy or possible. It allows us to find the rate of change of one variable with respect to another, no matter how entangled they are in the equation.