Math Problem Statement

Differentiate the following functions: A) y = (sin(x))^(1 + x^2) B) y = ln((x^2 + 1) / (x - 1))

Solution

We are tasked with differentiating two functions:

A) y=(sinx)1+x2y = (\sin x)^{1 + x^2}

B) y=ln(x2+1x1)y = \ln \left( \frac{x^2 + 1}{x - 1} \right)

I will solve each part step-by-step:


Part A: Differentiating y=(sinx)1+x2y = (\sin x)^{1 + x^2}

We use the logarithmic differentiation technique because the power includes a variable expression.

  1. Take the natural logarithm of both sides: lny=(1+x2)ln(sinx)\ln y = (1 + x^2) \ln (\sin x)

  2. Differentiate both sides with respect to xx, applying the product rule on the right-hand side: 1ydydx=ddx[(1+x2)ln(sinx)]\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ (1 + x^2) \ln (\sin x) \right]

    Using the product rule: [ \frac{d}{dx} \left[ (1 + x^2) \ln (\sin x) \right] = \frac{d}{dx} (1 + x^2) \cdot \ln (\sin x) + (1 + x^2) \cdot \frac{d}{dx} [\ln (\sin x)] ]

    • ddx(1+x2)=2x\frac{d}{dx} (1 + x^2) = 2x
    • ddx[ln(sinx)]=1sinxcosx=cotx\frac{d}{dx} [\ln (\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x

    So: 1ydydx=2xln(sinx)+(1+x2)cotx\frac{1}{y} \frac{dy}{dx} = 2x \ln (\sin x) + (1 + x^2) \cot x

  3. Multiply through by y=(sinx)1+x2y = (\sin x)^{1 + x^2} to isolate dydx\frac{dy}{dx}: dydx=(sinx)1+x2[2xln(sinx)+(1+x2)cotx]\frac{dy}{dx} = (\sin x)^{1 + x^2} \left[ 2x \ln (\sin x) + (1 + x^2) \cot x \right]


Part B: Differentiating y=ln(x2+1x1)y = \ln \left( \frac{x^2 + 1}{x - 1} \right)

Here we use the chain rule and the logarithmic property ln(uv)=lnulnv\ln \left( \frac{u}{v} \right) = \ln u - \ln v.

  1. Split the logarithm: y=ln(x2+1)ln(x1)y = \ln (x^2 + 1) - \ln (x - 1)

  2. Differentiate both terms separately:

    • For ln(x2+1)\ln (x^2 + 1), use the chain rule: ddx[ln(x2+1)]=1x2+12x=2xx2+1\frac{d}{dx} \left[ \ln (x^2 + 1) \right] = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}

    • For ln(x1)\ln (x - 1): ddx[ln(x1)]=1x1\frac{d}{dx} \left[ \ln (x - 1) \right] = \frac{1}{x - 1}

  3. Combine the results: dydx=2xx2+11x1\frac{dy}{dx} = \frac{2x}{x^2 + 1} - \frac{1}{x - 1}


Final Answers:

  1. A) dydx=(sinx)1+x2[2xln(sinx)+(1+x2)cotx]\frac{dy}{dx} = (\sin x)^{1 + x^2} \left[ 2x \ln (\sin x) + (1 + x^2) \cot x \right]
  2. B) dydx=2xx2+11x1\frac{dy}{dx} = \frac{2x}{x^2 + 1} - \frac{1}{x - 1}

Let me know if you'd like a detailed verification for any step or further clarification!

Related Questions:

  1. How do you differentiate functions with variable exponents using logarithmic differentiation?
  2. What is the product rule, and how does it apply to differentiation?
  3. Can you explain how to differentiate logarithmic functions like ln(uv)\ln \left( \frac{u}{v} \right)?
  4. How does the chain rule work in composite functions such as ( \ln (\sin x

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Math Problem Analysis

Mathematical Concepts

Logarithmic Differentiation
Chain Rule
Product Rule
Derivative of Logarithmic Functions

Formulas

d/dx [u^v] = u^v [v'ln(u) + v(u'/u)] (Logarithmic Differentiation)
d/dx [ln(u)] = u'/u
d/dx [uv] = u'v + uv' (Product Rule)

Theorems

Chain Rule
Logarithmic Properties: ln(a/b) = ln(a) - ln(b)

Suitable Grade Level

Undergraduate Calculus or Advanced High School (AP Calculus)