Question

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ h(y)=\tan ^{-1} y^{2} $$

Short answer

The critical point is \( y = 0 \), which is a local minimum.

Step-by-step solution

Find the First Derivative

To find the critical points, we first need to differentiate the function. Given the function \( h(y) = \tan^{-1}(y^2) \), let's find its derivative. We use the chain rule for the derivative of an inverse tangent function, \( \frac{d}{dy}[\tan^{-1}(u)] = \frac{1}{1+u^2} \cdot u' \). Thus, the derivative is:\[ h'(y) = \frac{1}{1+(y^2)^2} \cdot 2y = \frac{2y}{1+y^4} \]

Set the First Derivative to Zero

To find critical points, set the derivative \( h'(y) = \frac{2y}{1+y^4} \) equal to zero:\[ \frac{2y}{1+y^4} = 0 \]This equation equals zero when the numerator is zero. Therefore, we have \( 2y = 0 \) which means \( y = 0 \). Thus, \( y = 0 \) is a critical point.

Determine Sign Changes using the First Derivative Test

To use the First Derivative Test, consider points around \( y = 0 \), say \( y = -1 \) and \( y = 1 \). Substitute these points into the derivative:- For \( y = -1 \): \[ h'(-1) = \frac{2(-1)}{1+(-1)^4} = \frac{-2}{2} = -1 \] (negative)- For \( y = 1 \): \[ h'(1) = \frac{2(1)}{1+1^4} = \frac{2}{2} = 1 \] (positive)At \( y = 0 \), the derivative changes from negative to positive, indicating a local minimum.

Verify with the Second Derivative Test

Now, let's find the second derivative to use the Second Derivative Test. Differentiate the first derivative \( h'(y) = \frac{2y}{1+y^4} \) to get:\[ h''(y) = \frac{(2)(1+y^4) - (2y)(4y^3)}{(1+y^4)^2} \]Simplify:\[ h''(y) = \frac{2(1+y^4) - 8y^4}{(1+y^4)^2} = \frac{2 - 6y^4}{(1+y^4)^2} \]Evaluate \( h''(0) \):\[ h''(0) = \frac{2 - 6(0)^4}{(1+0^4)^2} = 2 \]Since \( h''(0) > 0 \), this confirms that \( y = 0 \) is a local minimum.

Key concepts

First Derivative Test

The First Derivative Test is a valuable tool in calculus for determining the local extremum (minimum or maximum) of a function at its critical points. Critical points occur where the derivative of a function is zero or undefined. Here, we started by finding the derivative of the given function, which was obtained as \( h'(y) = \frac{2y}{1+y^4} \). Setting this derivative to zero revealed the critical point at \( y = 0 \).
To apply the First Derivative Test, we evaluate the sign of the derivative on intervals around the critical point. We tested points \( y = -1 \) and \( y = 1 \), finding negative and positive signs, respectively. This sign change from negative to positive as we pass through \( y = 0 \) implies a local minimum, because the function descends towards the point and then ascends away from it.
This test is intuitive: when the derivative changes signs from negative to positive, the curve dips into a trough (a local minimum), and if it goes from positive to negative, it peaks into a crest (a local maximum). Nevertheless, remember that this test only works where the derivative actually changes signs.

Second Derivative Test

The Second Derivative Test provides another approach to confirming the nature of a critical point. Once you have identified a critical point by setting the first derivative to zero, as in the previous step, you can use the second derivative to determine if the point is a local minimum, maximum, or neither.
We found the second derivative of our function as \( h''(y) = \frac{2 - 6y^4}{(1+y^4)^2} \). Evaluating this at the critical point \( y = 0 \) yielded \( h''(0) = 2 \).
Since \( h''(0) > 0 \), the function is concave up at \( y = 0 \), meaning the point is a local minimum. If the second derivative had been negative, the function would be concave down, indicating a local maximum. If \( h''(y) \) had been zero, the test would be inconclusive, and further analysis would be needed.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \tan^{-1}(y^2) \), are used to find angles whose trigonometric ratios are known. These functions are significant in various calculus problems, especially when dealing with integrals and derivatives involving arc functions.
When differentiating inverse trigonometric functions, it's important to remember their basic derivative rules. The function \( \tan^{-1}(u) \) has a derivative formula of \( \frac{1}{1+u^2} \cdot u' \), where \( u \) is a function of another variable. This chain rule application allows us to effectively differentiate expressions involving inverse trigonometric functions.
In our example, \( h(y) = \tan^{-1}(y^2) \), the differentiation involved first recognizing \( y^2 \) as a nested function, applying the chain rule to derive the correct form for \( h'(y) \). Thus, mastering inverse trigonometric differentiation not only simplifies calculus problems but also broadens problem-solving strategies.

Chain Rule

The chain rule is a fundamental theorem in calculus used to differentiate compositions of functions. Simply put, if a function is a nest of more than one function, the chain rule gives us a way to tackle its derivative.
Consider the function \( h(y) = \tan^{-1}(y^2) \). This can be seen as a composition of \( \tan^{-1} \) and \( y^2 \). The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \times g'(x) \).
Applying the chain rule to our function, we differentiate \( \tan^{-1}(u) \) as mentioned in the previous sections: \( \frac{1}{1+u^2} \), where \( u = y^2 \). Then, we multiply by the derivative of \( y^2 \), which is \( 2y \), to give \( h'(y) = \frac{2y}{1+y^4} \).
The power of the chain rule is its ability to turn the calculus of compositional functions into a straightforward process. By breaking down complex derivatives into manageable steps, it becomes an indispensable tool in solving calculus problems efficiently.