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What are the differentials for the given functions and rules, including product rule and basic functions?

Mathematics

ScanSolve AI Answer

Answer

B

$2xdx$

Step-by-step solution

18 steps

1. If y=x2y = x^2 then dydy is:

  1. Differentiate yy with respect to xx.

    dydx=ddx(x2)=2x\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x
  2. Express the differential dydy.

    dy=2x dxdy = 2x \, dx

    The correct option is (B) 2xdx2xdx.

    2. If V=x3V = x^3, then differential of VV is:

  3. Differentiate VV with respect to xx.

    dVdx=ddx(x3)=3x2\frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2
  4. Express the differential dVdV.

    dV=3x2 dxdV = 3x^2 \, dx

    The correct option is (A) 3x2dx3x^2 dx.

    3. d(xy)=d(xy) =

  5. Apply the product rule for differentials, d(uv)=u dv+v dud(uv) = u \, dv + v \, du.

    Let u=xu = x and v=yv = y. Then du=dxdu = dx and dv=dydv = dy.

  6. Substitute into the product rule.

    d(xy)=x dy+y dxd(xy) = x \, dy + y \, dx

    The correct differential is x dy+y dx\boxed{x \, dy + y \, dx}.

    4. Differential of y\sqrt{y} is:

  7. Let u=y=y1/2u = \sqrt{y} = y^{1/2}.

  8. Differentiate uu with respect to yy.

    dudy=ddy(y1/2)=12y(1/2−1)=12y−1/2=12y\frac{du}{dy} = \frac{d}{dy}(y^{1/2}) = \frac{1}{2} y^{(1/2 - 1)} = \frac{1}{2} y^{-1/2} = \frac{1}{2\sqrt{y}}
  9. Express the differential dudu.

    du=12y dydu = \frac{1}{2\sqrt{y}} \, dy

    The correct differential is 12y dy\boxed{\frac{1}{2\sqrt{y}} \, dy}.

    5. Value of dydy, for y=x2y = x^2 and xx char:
    This question is incomplete as the full context for the change in xx is not provided.

    6. Differential of x\sqrt{x} is:

  10. Let y=x=x1/2y = \sqrt{x} = x^{1/2}.

  11. Differentiate yy with respect to xx.

    dydx=ddx(x1/2)=12x(1/2−1)=12x−1/2=12x\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{(1/2 - 1)} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
  12. Express the differential dydy.

    dy=12x dxdy = \frac{1}{2\sqrt{x}} \, dx

    The correct differential is 12x dx\boxed{\frac{1}{2\sqrt{x}} \, dx}.

    7. ∫sin⁡2x4sin⁡xdx=\int \frac{\sin 2x}{4 \sin x} dx =

  13. Use the double angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x.

    ∫2sin⁡xcos⁡x4sin⁡xdx\int \frac{2 \sin x \cos x}{4 \sin x} dx
  14. Simplify the integrand.

    ∫2cos⁡x4dx=∫12cos⁡x dx\int \frac{2 \cos x}{4} dx = \int \frac{1}{2} \cos x \, dx
  15. Integrate.

    12∫cos⁡x dx=12sin⁡x+C\frac{1}{2} \int \cos x \, dx = \frac{1}{2} \sin x + C

    The correct integral is 12sin⁡x+C\boxed{\frac{1}{2} \sin x + C}.

    8. ∫tan⁡π4dx\int \tan \frac{\pi}{4} dx

  16. Evaluate the constant tan⁡π4\tan \frac{\pi}{4}.

    tan⁡π4=1\tan \frac{\pi}{4} = 1
  17. Substitute the value into the integral.

    ∫1 dx\int 1 \, dx
  18. Integrate.

    ∫1 dx=x+C\int 1 \, dx = x + C

    The correct integral is x+C\boxed{x + C}.

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