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  1. J

    Finding more derivatives

    Homework Statement a.) f(x)=tan2(x) b.) cos3(x2) c.) (2x-1)/(5x+2) d.) (sqrt(x2-2x))(secx) e.) f(x)=((2x+3)/(x+7))3/2 f.) [sin(x)cos(x)]2 Homework Equations chain rule Product rule Quotient rule Power rule The Attempt at a Solution a.) would you do the power rule...
  2. J

    Finding the derivative

    (x2-x)cosx+sinx(2x-1) X2cosx-xcosx+2xsinx-sinx
  3. J

    Finding the derivative

    Homework Statement f(x)=(x2-x)(sin(x)) Homework Equations Chain rule Product rule The Attempt at a Solution f'g(x)*g'(x) cosx (2x-x)(cosx) Would you use the product rule after using the power rule and chain rule?
  4. J

    Rotation of a Rigid Object About a Fixed Axis

    Homework Statement A small object with mass 4.00kg moves counterclockwise with constant speed 4.50m/s in a circle of radius 3.00m centered at the origin. It starts at the point with position vector (3.00i+0j)m. Then it undergoes an angular displacement of 9.00rad. (a.) What it is position...
  5. J

    Finding the Derivative

    x=2 [2(-3)-4(0)]/2^2 =-3/2
  6. J

    Finding the Derivative

    \frac{x[h'(x)]-h(x)x'}{x^{2}}
  7. J

    Finding the Derivative

    Um, h(x)=4, because x=2 so h(2)=4; x=2, so I get 1/2. But you don't understand how you would the the quotient rule using that value. There is the product rule if you rearrange the formula to 1(2)^-1 I'm sorry, I'm a little confused.
  8. J

    Find Derivatives of a Function

    16! Thank you very much :)
  9. J

    Find Derivatives of a Function

    f'(x)=X^(1/2)g'(x)+g(x)(1/2)x^(-1/2) ?
  10. J

    Find Derivatives of a Function

    f'(x)=X^(1/2)g'g(x)+g(x)(1/2)x^(-1/2) ?
  11. J

    Finding the Derivative

    SO the derivative would be -3/4?
  12. J

    Find Derivatives of a Function

    f'(x)=x^{\frac{1}{2}}7+8(\frac{1}{2}x^{-\frac{1}{2}})
  13. J

    Find Derivatives of a Function

    f(x)=x^{\frac{1}{2}}g(x) f'(x)=\frac{1}{2}x^{-\frac{1}{2}}g(x)
  14. J

    Find Derivatives of a Function

    The product rule is d/dx [f(x)*g(x)]=f(x)g'(x)+g(x)f'(x)
  15. J

    Finding the Derivative

    Homework Statement If h(2)=4 and h'(2)=-3, find \left.\frac{d}{dx}\frac{h(x)}{x}\right|_{x=2} Homework Equations n^n-1 (power rule) The Attempt at a Solution I don't know how to get this started. It seems like I am having trouble with derivatives. I can do simple derivatives with...
  16. J

    Find Derivatives of a Function

    Homework Statement If f(x)=\sqrt{x}g(x), where g(4)=8 and g'(4)=7, find f'(4). Homework Equations the product rule/quotient rule The Attempt at a Solution I am having trouble getting this problem started. Would I solve for the x of f(x)?
  17. J

    Kinetic Energy and the Work-Kinetic Energy

    What I did is Loss in energy=Loss in PE from the topmost point=mg(5.12) Avg. Force=mg(5.12)/0.12 =8.78x10^5N
  18. J

    Cross products vs. dot products

    So what you are trying to convey is that you multiply with cross products and dot products, but it is WHAT you multiply determines whether it is a dot or cross product? Scalar is always dot and vector is always cross product?
  19. J

    Kinetic Energy and the Work-Kinetic Energy

    Homework Statement A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam. Then it drives the beam 12.0 cm farther into the ground as it comes to rest. Using energy considerations, calculate the...
  20. J

    Cross products vs. dot products

    I have a question regarding cross products and dot products. What is the difference, and are there any similarities? What ARE cross products and what are their functions? What ARE dot products and what are their functions? What do we use them for? Thank you, Joyci116
  21. J

    Collisions in One Dimension

    Thanks, I finally figured it out. :D
  22. J

    Collisions in One Dimension

    p=mv The momentum found using final velocity: p=(0.01kg)(0.600m/s)=0.006 kg*m/s (ball) p=(5.00kg)(0.600m/s)=0.006 kg*m/s (wood) And I assumed the initial velocity was zero, so I ended up with a zero momentum. Is that correct so far?
  23. J

    Collisions in One Dimension

    Homework Statement A 10.0g bullet is fired into a stationary block of wood (m=5.00 kg). The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet? ball m=10.0g=0.01kg wood m=...
  24. J

    Momentum and Kinetic Energy

    Thank you for the help. :)
  25. J

    Momentum and Kinetic Energy

    K=p^2/2m K=.5mV^2 p=mv V=p/m K=.5m(p/m)^2 K=p^2/2m
  26. J

    Momentum and Kinetic Energy

    The equation for kinetic energy is k=.5mV^2 and the new equation is K=p^2/2m. Momentum is missing from the first equation and velocity is missing from the second equation.
  27. J

    Momentum and Kinetic Energy

    Homework Statement (a) A particle of mass m moves with momentum of magnitude p. Show that the kinetic energy of the particle is given by K=p^{2}/2m. (b) Express the magnitude of the particle's momentum in terms of its kinetic energy and mass. Homework Equations Would the p=mv be needed...
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