Math Problem Statement

If y equals open parentheses cos space x close parentheses to the power of x squared end exponent comma spacethen fraction numerator d y over denominator d x end fraction equals

a. open parentheses cos space x close parentheses to the power of x squared end exponent open parentheses negative 2 x space tan space x space plus x squared space ln left parenthesis cos x right parenthesis close parentheses

b. x squared space open parentheses cos space x close parentheses to the power of x squared minus 1 end exponent open parentheses negative sin space x close parentheses left parenthesis 2 x right parenthesis

c. negative x squared space open parentheses cos space x close parentheses to the power of x squared end exponent open parentheses tan space x space minus ln left parenthesis cos space x right parenthesis close parentheses

d. negative x space open parentheses cos space x close parentheses to the power of x squared end exponent open parentheses x space tan space x space minus 2 space ln left parenthesis cos space x right parenthesis close parentheses

e. negative x space open parentheses cos space x close parentheses to the power of x squared end exponent open parentheses x space tan space x space plus 2 space ln left parenthesis cos space x right parenthesis close parentheses

Solution

To find dydx\frac{dy}{dx} where y=(cosx)x2y = (\cos x)^{x^2}, we will use logarithmic differentiation.

Given: y=(cosx)x2y = (\cos x)^{x^2}

  1. Take the natural logarithm of both sides: lny=ln((cosx)x2)\ln y = \ln \left( (\cos x)^{x^2} \right) lny=x2ln(cosx)\ln y = x^2 \ln (\cos x)

  2. Differentiate both sides with respect to xx: ddx(lny)=ddx(x2ln(cosx))\frac{d}{dx} (\ln y) = \frac{d}{dx} (x^2 \ln (\cos x))

Using the chain rule on the left side and the product rule on the right side: 1ydydx=2xln(cosx)+x2ddx(ln(cosx))\frac{1}{y} \frac{dy}{dx} = 2x \ln (\cos x) + x^2 \cdot \frac{d}{dx} (\ln (\cos x))

  1. The derivative of ln(cosx)\ln (\cos x) is: ddx(ln(cosx))=1cosx(sinx)=tanx\frac{d}{dx} (\ln (\cos x)) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x

Thus, 1ydydx=2xln(cosx)+x2(tanx)\frac{1}{y} \frac{dy}{dx} = 2x \ln (\cos x) + x^2 (-\tan x) 1ydydx=2xln(cosx)x2tanx\frac{1}{y} \frac{dy}{dx} = 2x \ln (\cos x) - x^2 \tan x

  1. Multiply both sides by yy to solve for dydx\frac{dy}{dx}: dydx=y(2xln(cosx)x2tanx)\frac{dy}{dx} = y \left( 2x \ln (\cos x) - x^2 \tan x \right)

Substitute y=(cosx)x2y = (\cos x)^{x^2}: dydx=(cosx)x2(2xln(cosx)x2tanx)\frac{dy}{dx} = (\cos x)^{x^2} \left( 2x \ln (\cos x) - x^2 \tan x \right)

Comparing this with the given options, we have:

a. (cosx)x2(2xtanx+x2ln(cosx))(\cos x)^{x^2} (-2x \tan x + x^2 \ln (\cos x))

b. x2(cosx)x21(sinx)(2x)x^2 (\cos x)^{x^2-1} (-\sin x) (2x)

c. x2(cosx)x2(tanxln(cosx))-x^2 (\cos x)^{x^2} (\tan x - \ln (\cos x))

d. x(cosx)x2(xtanx2ln(cosx))-x (\cos x)^{x^2} (x \tan x - 2 \ln (\cos x))

e. x(cosx)x2(xtanx+2ln(cosx))-x (\cos x)^{x^2} (x \tan x + 2 \ln (\cos x))

The correct answer matches option (a).

a\boxed{a}

Would you like more details or have any questions? Here are 5 more questions you can ask:

  1. Can you explain the product rule in calculus?
  2. What is the chain rule and how is it used in differentiation?
  3. How do you find the derivative of a function raised to a power?
  4. What is logarithmic differentiation and when is it used?
  5. Can you solve another example involving logarithmic differentiation?

Tip: When solving complex differentiation problems, breaking them down using rules like the chain rule, product rule, and logarithmic differentiation can simplify the process and reduce errors.

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Math Problem Analysis

Mathematical Concepts

Logarithmic differentiation
Chain rule
Product rule

Formulas

\( \frac{d}{dx} (\ln u) = \frac{1}{u} \frac{du}{dx} \)
\( \frac{d}{dx} (uv) = u'v + uv' \)

Theorems

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Suitable Grade Level

Advanced Mathematics