Math Problem Statement

Solve the ten problems

Solve the following problems. Differentiate the functions from the image: (1) y = x^2 - 2x^4 - 3x^5, (2) y = 5x^{-2} + x^{-3}, (3) y = 1/v^2 - 2/v^4, (4) y = 2/3 - y^2/5 + 4y^3/3, (5) y = 6 - x^{-4/3}, (6) x = sqrt(t) - 1/sqrt(t), (7) y = (v^2 + 3)(v^2 - v + 1), (8) x = 3/(t^2 - 1), (9) y = x/(x^2 - 1), (10) f(x) = x(x+1)/(x^2 - 1).

Solution

Let's go through the process of differentiating each function one by one.

1. y=x22x43x5y = x^2 - 2x^4 - 3x^5

Differentiate term by term using the power rule: ddxxn=nxn1\frac{d}{dx} x^n = nx^{n-1}.

dydx=2x8x315x4\frac{dy}{dx} = 2x - 8x^3 - 15x^4

2. y=5x2+x3y = 5x^{-2} + x^{-3}

Again, apply the power rule.

dydx=10x33x4\frac{dy}{dx} = -10x^{-3} - 3x^{-4}

3. y=1v22v4y = \frac{1}{v^2} - \frac{2}{v^4}

Rewrite the terms as powers of vv and then differentiate.

y=v22v4y = v^{-2} - 2v^{-4} dydv=2v3+8v5\frac{dy}{dv} = -2v^{-3} + 8v^{-5}

4. y=23y25+4y33y = \frac{2}{3} - \frac{y^2}{5} + \frac{4y^3}{3}

Differentiate term by term. The constant 23\frac{2}{3} will disappear since the derivative of a constant is 0.

dydy=02y5+4y2\frac{dy}{dy} = 0 - \frac{2y}{5} + 4y^2

5. y=6x4/3y = 6 - x^{-4/3}

Differentiate both terms. The constant 6 disappears, and apply the power rule to the second term.

dydx=43x7/3\frac{dy}{dx} = \frac{4}{3}x^{-7/3}

6. x=t1tx = \sqrt{t} - \frac{1}{\sqrt{t}}

Rewrite in terms of powers of tt, i.e., t1/2t^{1/2} and t1/2t^{-1/2}, then differentiate.

dxdt=12t1/2+12t3/2\frac{dx}{dt} = \frac{1}{2}t^{-1/2} + \frac{1}{2}t^{-3/2}

7. y=(v2+3)(v2v+1)y = (v^2 + 3)(v^2 - v + 1)

Use the product rule: ddv(uv)=uv+uv\frac{d}{dv}(uv) = u'v + uv'.

dydv=(2v)(v2v+1)+(v2+3)(2v1)\frac{dy}{dv} = \left(2v\right)(v^2 - v + 1) + (v^2 + 3)(2v - 1) Simplify: dydv=2v(v2v+1)+(v2+3)(2v1)\frac{dy}{dv} = 2v(v^2 - v + 1) + (v^2 + 3)(2v - 1)

8. x=3t21x = \frac{3}{t^2 - 1}

This is a quotient. Use the quotient rule: ddt(uv)=uvuvv2\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

dxdt=0(t21)3(2t)(t21)2=6t(t21)2\frac{dx}{dt} = \frac{0(t^2 - 1) - 3(2t)}{(t^2 - 1)^2} = \frac{-6t}{(t^2 - 1)^2}

9. y=xx21y = \frac{x}{x^2 - 1}

Use the quotient rule again.

dydx=(1)(x21)x(2x)(x21)2=x212x2(x21)2=x21(x21)2\frac{dy}{dx} = \frac{(1)(x^2 - 1) - x(2x)}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2}

10. f(x)=x(x+1)x21f(x) = \frac{x(x+1)}{x^2 - 1}

Use the product rule and the quotient rule together.

First, differentiate x(x+1)=x2+xx(x+1) = x^2 + x, which gives 2x+12x + 1.

Now use the quotient rule: f(x)=(2x+1)(x21)(x2+x)(2x)(x21)2f'(x) = \frac{(2x+1)(x^2 - 1) - (x^2 + x)(2x)}{(x^2 - 1)^2} Simplify this expression.


Do you need any of the solutions simplified further or explained in more detail? Feel free to ask for clarification!

Related Questions:

  1. What is the power rule for differentiation and how is it applied?
  2. How do you handle differentiation of products like in problem 7?
  3. What is the quotient rule and how does it work for problems 8 and 9?
  4. Why do constants vanish when differentiating?
  5. What are common mistakes to avoid when using the product and quotient rules?

Tip:

Always simplify the function as much as possible before differentiating to avoid unnecessary complications!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Power Rule
Quotient Rule
Product Rule

Formulas

Power Rule: d/dx(x^n) = n*x^{n-1}
Quotient Rule: d/dx(u/v) = (u'v - uv')/v^2
Product Rule: d/dx(uv) = u'v + uv'

Theorems

Power Rule Theorem
Quotient Rule Theorem
Product Rule Theorem

Suitable Grade Level

Grades 11-12 or Early University Calculus