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Chapter 3: Problem 19

In Exercises \(13-22,\) find the derivatives of \(y\) with respect to the appropriate variable. $$y=\cot ^{-1} \sqrt{t-1}$$

Short Answer

Expert verified
The derivative of \(y\) with respect to \(t\) is \(dy/dt = -1/(2\sqrt{t-1} * t)\).

Step by step solution

01

Identify The Function And Derivative Formulas

The function is \(y=\cot ^{-1} \sqrt{t-1}\). Here, it needs to recognize that this function is a composition of the inverse cotangent function and the square root function. The derivative of \(arccot (x)\) is \(-1/(1+x^2)\). Also, the derivative of \(\sqrt {x}\) is \(1/(2\sqrt{x})\).
02

Apply the Chain Rule

The chain rule states that the derivative of a composition of functions is the derivative of the outer function multiplied by the derivative of the inner function. Symbolically, if \(y = f(g(x))\), then \(dy/dx = f'(g(x)) * g'(x)\). Applying chain rule here, we first look at \(\cot^{-1}\) as the 'outside' function and \(\sqrt{t-1}\) as the 'inside' function. The derivative of \(y\) with respect to \(t\) becomes: \(dy/dt = -1/(1+(\sqrt{t-1})^2) * d/dt (\sqrt{t-1})\).
03

Find Derivative of the Inner Function

The derivative of the square root function can be found after rewriting it with an exponent of \(1/2\). It becomes, \(d/dt (\sqrt{t-1}) = d/dt ((t-1)^{1/2}) = 1/2 * (t-1)^{-1/2} = 1/(2\sqrt{t-1})\).
04

Substitute the Derivative of the Inner Function

Substitute \(d/dt (\sqrt{t-1})\) into the formula derived in step 2. \(dy/dt\) becomes \(-1/ (1+(\sqrt{t-1})^2) * 1/(2\sqrt{t-1}) = -1/(2\sqrt{t-1} * (1+(t-1)))\).
05

Simplify if Possible

Simplifying the expression, the derivative \(dy/dt = -1/(2\sqrt{t-1} * t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are essential components in calculus. They essentially "undo" the trigonometric functions, like sine, cosine, and tangent. In our exercise, we are dealing with the inverse cotangent function, written as \( \cot^{-1}(x) \). This function flips what the original trigonometric function does. Instead of outputting a ratio, it outputs an angle.

To derive the inverse trig functions, it's critical to know their formulas. Here are a few things to remember about \( \cot^{-1}(x) \):
  • The derivative is \(-1/(1+x^2)\), a result of its unique characteristics.
  • This differs from the derivative of \( \tan^{-1}(x) \), which is \(1/(1+x^2)\).
  • Erroneously mixing these formulas could lead to mistakes in calculations.
Inverse trigonometric functions are part of the tools for taming complex expressions while differentiating, especially in calculus problems that include nested functions.
Chain Rule
The chain rule is a powerful tool in calculus for finding the derivative of a composite function. A composite function is one where one function is applied to the result of another, like \( f(g(x)) \). In our problem, the function is \( y = \cot^{-1}(\sqrt{t-1}) \).

Here鈥檚 the basic idea:
  • Identify the "outer" function, which in this case is \( \cot^{-1}() \).
  • Recognize the "inner" function, which here is \( \sqrt{t-1} \).
  • Find the derivative of each function independently.
  • Multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function.
When using the chain rule, remember that each component must be differentiated separately, then combined.
The key to mastering the chain rule is practice and recognizing the structure of composite functions.
Derivative of Composite Functions
Taking the derivative of composite functions can initially seem daunting, yet the process simplifies with the use of the chain rule. For the exercise \( y = \cot^{-1}(\sqrt{t-1}) \), both functions need individual attention: the inverse trig function and the square root.

Understanding their derivatives:
  • The outer function, \( \cot^{-1}(x) \), differentiates to \(-1/(1+x^2)\).
  • The inner function, \( \sqrt{t-1} \), simplifies to an exponent: \((t-1)^{1/2}\).
  • Derivate \( (t-1)^{1/2} \) to get \(1/(2\sqrt{t-1})\).
As you combine these derivatives using the chain rule, you鈥檙e left with a simplified expression.
Despite initial complexity, breaking down each step and applying the chain rule can make differentiation manageable and intuitive.

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Most popular questions from this chapter

At what point on the graph of \(y=3^{x}+1\) is the tangent line parallel to the line \(y=5 x-1 ?\)

Radians vs. Degrees What happens to the derivatives of \(\sin x\) and cos \(x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. (a) With your grapher in degree mode, graph \(f(h)=\frac{\sin h}{h}\) and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) (b) With your grapher in degree mode, estimate \(\lim _{h \rightarrow 0} \frac{\cos h-1}{h}\) (c) Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? (d) Derive the formula for the derivative of cos \(x\) using degree-mode limits. (e) The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. What are the second and third degree-mode derivatives of \(\sin x\) and \(\cos x\) ?

Orthogonal Families of Curves Prove that all curves in the family \(y=-\frac{1}{2} x^{2}+k\)

Identities Confirm the following identities for \(x>0\) . (a) \(\cos ^{-1} x+\sin ^{-1} x=\pi / 2\) (b) \(\tan ^{-1} x+\cot ^{-1} x=\pi / 2\) (c) \(\sec ^{-1} x+\csc ^{-1} x=\pi / 2\)

Standardized Test Questions You should solve the following problems without using a graphing calculator. True or False The derivative of \(y=2^{x}\) is \(2^{x} .\) Justify your answer.
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