Math Problem Statement

Find yn, if y = tan^−1 (x/c)

Solution

To find yny_n (the nth derivative of yy) if y=tan1(xc)y = \tan^{-1}\left(\frac{x}{c}\right), let's follow these steps.

Step 1: Initial function

We are given: y=tan1(xc)y = \tan^{-1}\left(\frac{x}{c}\right)

Step 2: First derivative y1y_1

The derivative of tan1(u)\tan^{-1}(u) with respect to uu is: ddutan1(u)=11+u2\frac{d}{du} \tan^{-1}(u) = \frac{1}{1 + u^2} Now, apply the chain rule. Let u=xcu = \frac{x}{c}, then: y1=ddx(tan1(xc))=11+(xc)2ddx(xc)y_1 = \frac{d}{dx} \left( \tan^{-1}\left(\frac{x}{c}\right) \right) = \frac{1}{1 + \left(\frac{x}{c}\right)^2} \cdot \frac{d}{dx} \left( \frac{x}{c} \right) Since ddx(xc)=1c\frac{d}{dx} \left( \frac{x}{c} \right) = \frac{1}{c}, we get: y1=1c11+(xc)2=1c11+x2c2y_1 = \frac{1}{c} \cdot \frac{1}{1 + \left(\frac{x}{c}\right)^2} = \frac{1}{c} \cdot \frac{1}{1 + \frac{x^2}{c^2}} Simplifying: y1=1cc2c2+x2=cc2+x2y_1 = \frac{1}{c} \cdot \frac{c^2}{c^2 + x^2} = \frac{c}{c^2 + x^2}

Step 3: Second derivative y2y_2

Now, find the second derivative by differentiating y1y_1: y2=ddx(cc2+x2)y_2 = \frac{d}{dx} \left( \frac{c}{c^2 + x^2} \right) Using the quotient rule, ddx(f(x)g(x))=f(x)g(x)f(x)g(x)g(x)2\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}, where:

  • f(x)=cf(x) = c, so f(x)=0f'(x) = 0
  • g(x)=c2+x2g(x) = c^2 + x^2, so g(x)=2xg'(x) = 2x

Applying the quotient rule: y2=0(c2+x2)c2x(c2+x2)2=2cx(c2+x2)2y_2 = \frac{0 \cdot (c^2 + x^2) - c \cdot 2x}{(c^2 + x^2)^2} = \frac{-2cx}{(c^2 + x^2)^2}

Step 4: Higher derivatives yny_n

We can see a pattern emerging in the derivatives. The first derivative y1y_1 is positive, and the second derivative y2y_2 introduces a negative factor with xx in the numerator. We can generalize that:

  • Odd derivatives will have terms involving xx raised to odd powers and will alternate in sign.
  • Even derivatives will have terms involving xx raised to even powers, also alternating in sign.

For higher-order derivatives yny_n, the pattern involves successively differentiating using the quotient rule. Each differentiation introduces higher powers of xx in the numerator and increases the power of (c2+x2)(c^2 + x^2) in the denominator.

Would you like to continue with more specific higher-order derivatives or explore this pattern in more depth?

Related Questions:

  1. How is the inverse tangent function derived?
  2. What is the geometric interpretation of y=tan1(x/c)y = \tan^{-1}(x/c)?
  3. Can we use implicit differentiation to find higher derivatives of inverse trigonometric functions?
  4. How does the behavior of y=tan1(x/c)y = \tan^{-1}(x/c) change as xx \to \infty?
  5. What are some applications of inverse trigonometric functions in calculus?

Tip:

When differentiating functions involving inverse trigonometric expressions, always check for simplifications to reduce the complexity of higher derivatives.

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Math Problem Analysis

Mathematical Concepts

Calculus
Inverse Trigonometric Functions
Differentiation

Formulas

y = tan^−1(x/c)
dy/dx = 1 / (1 + u^2)
Quotient rule: (f'(x)g(x) - f(x)g'(x)) / g(x)^2

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

College level (Calculus I/II)