Math Problem Statement

Find the derivatives as indicated in the following: a) g(x) = 3x^2 + 4 ; g'(1) b) h(r) = r^2 + 2^r ; h'(r) c) z = 9^{4 log_9(x)} ; z'' d) y(x) = sqrt(e^{2x} + ln(x^2)) ; y'(x)

Solution

Let's solve each part step-by-step.

Part (a)

Given: g(x)=3x2+4g(x) = 3x^2 + 4 We are asked to find g(1)g'(1).

  1. Differentiate g(x)g(x) with respect to xx: g(x)=ddx(3x2+4)=32x=6xg'(x) = \frac{d}{dx}(3x^2 + 4) = 3 \cdot 2x = 6x

  2. Evaluate g(1)g'(1): g(1)=61=6g'(1) = 6 \cdot 1 = 6

Answer: g(1)=6g'(1) = 6.


Part (b)

Given: h(r)=r2+2rh(r) = r^2 + 2^r We are asked to find h(r)h'(r).

  1. Differentiate h(r)h(r) with respect to rr:

    • For r2r^2, the derivative is 2r2r.
    • For 2r2^r, we use the rule for exponential functions: ddr(ar)=arln(a)\frac{d}{dr}(a^r) = a^r \ln(a).

    Thus: h(r)=2r+2rln(2)h'(r) = 2r + 2^r \ln(2)

Answer: h(r)=2r+2rln(2)h'(r) = 2r + 2^r \ln(2).


Part (c)

Given: z=94log9(x)z = 9^{4 \log_9(x)} We are asked to find zz'', the second derivative of zz with respect to xx.

  1. Simplify zz: Using properties of exponents and logarithms, 94log9(x)9^{4 \log_9(x)} simplifies as follows: z=x4z = x^4

  2. Find the first derivative, zz': z=ddx(x4)=4x3z' = \frac{d}{dx}(x^4) = 4x^3

  3. Find the second derivative, zz'': z=ddx(4x3)=12x2z'' = \frac{d}{dx}(4x^3) = 12x^2

Answer: z=12x2z'' = 12x^2.


Part (d)

Given: y(x)=e2x+ln(x2)y(x) = \sqrt{e^{2x} + \ln(x^2)} We are asked to find y(x)y'(x).

  1. Rewrite y(x)y(x) for easier differentiation: y(x)=(e2x+ln(x2))1/2y(x) = \left(e^{2x} + \ln(x^2)\right)^{1/2}

  2. Differentiate y(x)y(x) using the chain rule: Let u=e2x+ln(x2)u = e^{2x} + \ln(x^2), so y=u1/2y = u^{1/2}. Then, y=12u1/2uy' = \frac{1}{2} u^{-1/2} \cdot u'

    Now, find uu':

    • For e2xe^{2x}, the derivative is 2e2x2e^{2x}.
    • For ln(x2)\ln(x^2), use the chain rule: ddxln(x2)=1x22x=2x\frac{d}{dx} \ln(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}.

    So, u=2e2x+2xu' = 2e^{2x} + \frac{2}{x}

  3. Combine the results: y=121e2x+ln(x2)(2e2x+2x)y' = \frac{1}{2} \cdot \frac{1}{\sqrt{e^{2x} + \ln(x^2)}} \cdot \left(2e^{2x} + \frac{2}{x}\right) Simplifying further: y=e2x+1xe2x+ln(x2)y' = \frac{e^{2x} + \frac{1}{x}}{\sqrt{e^{2x} + \ln(x^2)}}

Answer: y=e2x+1xe2x+ln(x2)y' = \frac{e^{2x} + \frac{1}{x}}{\sqrt{e^{2x} + \ln(x^2)}}.


Would you like further details on any of these solutions, or do you have specific questions? Here are some related questions to expand on these concepts:

  1. How do you find higher-order derivatives of polynomial functions?
  2. What is the chain rule, and how is it applied in derivative calculations?
  3. How can we simplify expressions involving logarithmic and exponential functions before differentiating?
  4. Why do exponential functions of the form axa^x have derivatives involving ln(a)\ln(a)?
  5. How does implicit differentiation work, and when would you use it?

Tip: When differentiating complex functions, try simplifying the expression first—it often makes the differentiation process easier.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Exponential and Logarithmic Functions
Chain Rule
Power Rule
Logarithmic Properties

Formulas

Power Rule: d/dx(x^n) = nx^(n-1)
Exponential Derivative: d/dx(a^x) = a^x ln(a)
Logarithmic Derivative: d/dx(log_a(x)) = 1/(x ln(a))
Chain Rule: (f(g(x)))' = f'(g(x)) * g'(x)

Theorems

Properties of Exponents and Logarithms

Suitable Grade Level

College Level Calculus