Math Problem Statement

Find dy/dx for the following functions: (a) f(x) = 10^(2x-x^2) + x ln x^2, (b) x = t^2 + 4t, y = tan t, (c) y = log(u^2 - 2), u = sqrt(x), (d) If g(x) = f^(-1)(x), where f(x) = e^(2x).

Solution

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(a) f(x)=102xx2+xlnx2f(x) = 10^{2x - x^2} + x \ln x^2

  1. Use the chain rule and logarithmic differentiation to find ddx\frac{d}{dx}.
    • First term: ddx[102xx2]=102xx2ln(10)ddx(2xx2)=102xx2ln(10)(22x)\frac{d}{dx}[10^{2x - x^2}] = 10^{2x - x^2} \cdot \ln(10) \cdot \frac{d}{dx}(2x - x^2) = 10^{2x - x^2} \cdot \ln(10) \cdot (2 - 2x).
    • Second term: ddx[xlnx2]=ddx[x2lnx]=ddx[2xlnx]=2lnx+2\frac{d}{dx}[x \ln x^2] = \frac{d}{dx}[x \cdot 2 \ln x] = \frac{d}{dx}[2x \ln x] = 2 \ln x + 2.

Final Answer: dfdx=102xx2ln(10)(22x)+2lnx+2\frac{df}{dx} = 10^{2x - x^2} \ln(10)(2 - 2x) + 2 \ln x + 2


(b) Parametric differentiation: x=t2+4tx = t^2 + 4t, y=tanty = \tan t

  1. Differentiate xx and yy with respect to tt:

    • dxdt=2t+4\frac{dx}{dt} = 2t + 4
    • dydt=sec2t\frac{dy}{dt} = \sec^2 t
  2. Use dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}:

    • dydx=sec2t2t+4\frac{dy}{dx} = \frac{\sec^2 t}{2t + 4}.

Final Answer: dydx=sec2t2t+4\frac{dy}{dx} = \frac{\sec^2 t}{2t + 4}


(c) y=log(u22)y = \log(u^2 - 2), u=xu = \sqrt{x}

  1. Substitute u=xu = \sqrt{x}: y=log((x)22)=log(x2)y = \log((\sqrt{x})^2 - 2) = \log(x - 2).

  2. Differentiate yy with respect to xx:

    • dydx=1x2ddx[x2]=1x2\frac{dy}{dx} = \frac{1}{x - 2} \cdot \frac{d}{dx}[x - 2] = \frac{1}{x - 2}.

Final Answer: dydx=1x2\frac{dy}{dx} = \frac{1}{x - 2}


(d) g(x)=f1(x)g(x) = f^{-1}(x), where f(x)=e2xf(x) = e^{2x}

  1. For inverse functions: f(g(x))=xf(g(x)) = x.

    • Differentiate both sides: f(g(x))g(x)=1f'(g(x)) \cdot g'(x) = 1.
  2. Compute f(x)=ddx[e2x]=2e2xf'(x) = \frac{d}{dx}[e^{2x}] = 2e^{2x}.

  3. Substituting f(g(x))=2e2g(x)f'(g(x)) = 2e^{2g(x)}:

    • g(x)=12e2g(x)g'(x) = \frac{1}{2e^{2g(x)}}.
  4. Since g(x)g(x) is the inverse, g(x)=12ln(x)g(x) = \frac{1}{2} \ln(x):

    • g(x)=12xg'(x) = \frac{1}{2x}.

Final Answer: g(x)=12xg'(x) = \frac{1}{2x}


Let me know if you'd like further explanations or examples. Here are 5 questions for deeper exploration:

  1. How does the chain rule apply to logarithmic differentiation in part (a)?
  2. Why is parametric differentiation necessary in part (b)?
  3. Could you generalize part (c) for uu being any function of xx?
  4. How does the derivative of an inverse function relate to its original function in part (d)?
  5. What are the steps to simplify expressions with logarithmic or exponential functions?

Tip: Always rewrite functions in a simpler form before differentiating to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Logarithmic Differentiation
Chain Rule
Parametric Differentiation
Derivative of Logarithmic and Exponential Functions
Inverse Functions and Their Derivatives

Formulas

d/dx [a^u] = a^u ln(a) du/dx
d/dx [ln(x)] = 1/x
d/dx [log_b(x)] = 1/(x ln(b))
dy/dx = (dy/dt) / (dx/dt)
Inverse Function Derivative: g'(x) = 1 / f'(g(x))

Theorems

Chain Rule
Properties of Logarithms
Inverse Function Theorem

Suitable Grade Level

Undergraduate