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Question 75 of 99

Q.The differential of yy if y=x4+x2+1y = \sqrt{x^4 + x^2 + 1} is :

(a) 12(4x3+2x)−12\dfrac{1}{2}(4x^3 + 2x)^{-\frac{1}{2}}
(b) 12(4x3+2x)−12 dx\dfrac{1}{2}(4x^3 + 2x)^{-\frac{1}{2}}\,dx
(c) 12(x4+x2+1)−12(4x3+2x)\dfrac{1}{2}(x^4 + x^2 + 1)^{-\frac{1}{2}}(4x^3 + 2x)
(d) 12(x4+x2+1)−12(4x3+2x) dx\dfrac{1}{2}(x^4 + x^2 + 1)^{-\frac{1}{2}}(4x^3 + 2x)\,dx
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Applying the chain rule to y=x4+x2+1y=\sqrt{x^4+x^2+1} and multiplying by dxdx gives the differential dy=12(x4+x2+1)−1/2(4x3+2x) dxdy=\tfrac12(x^4+x^2+1)^{-1/2}(4x^3+2x)\,dx.

  1. Write y=(x4+x2+1)1/2y=(x^4+x^2+1)^{1/2}.
  2. By the chain rule, dydx=12(x4+x2+1)−12⋅ddx(x4+x2+1)\dfrac{dy}{dx} = \dfrac12(x^4+x^2+1)^{-\frac12}\cdot\dfrac{d}{dx}(x^4+x^2+1).
  3. Compute the inner derivative: ddx(x4+x2+1)=4x3+2x\dfrac{d}{dx}(x^4+x^2+1) = 4x^3+2x.
  4. So dydx=12(x4+x2+1)−12(4x3+2x)\dfrac{dy}{dx} = \dfrac12(x^4+x^2+1)^{-\frac12}(4x^3+2x). …

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