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Find the derivative of x with respect to y for the equation x = (3/4)cot(2y - pi/2).

Mathematics

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Answer

−32csc⁡2(2y−π2)-\frac{3}{2} \csc^2 \left(2y - \frac{\pi}{2}\right)

Step-by-step solution

5 steps

To find dxdy\frac{dx}{dy} for the given equation x=34cot⁡(2y−π2)x = \frac{3}{4} \cot \left(2y - \frac{\pi}{2}\right), we will use the chain rule.

  1. Identify the outer and inner functions.

    Let the outer function be f(u)=34cot⁡uf(u) = \frac{3}{4} \cot u and the inner function be u=2y−π2u = 2y - \frac{\pi}{2}.

  2. Differentiate the inner function with respect to yy.

    dudy=ddy(2y−π2)=2\frac{du}{dy} = \frac{d}{dy}\left(2y - \frac{\pi}{2}\right) = 2

  3. Differentiate the outer function with respect to uu.

    Recall that ddu(cot⁡u)=−csc⁡2u\frac{d}{du}(\cot u) = -\csc^2 u.

    dxdu=ddu(34cot⁡u)=34(−csc⁡2u)=−34csc⁡2u\frac{dx}{du} = \frac{d}{du}\left(\frac{3}{4} \cot u\right) = \frac{3}{4} (-\csc^2 u) = -\frac{3}{4} \csc^2 u

  4. Apply the chain rule, dxdy=dxdu⋅dudy\frac{dx}{dy} = \frac{dx}{du} \cdot \frac{du}{dy}.

    dxdy=(−34csc⁡2u)⋅(2)\frac{dx}{dy} = \left(-\frac{3}{4} \csc^2 u\right) \cdot (2)

  5. Substitute u=2y−π2u = 2y - \frac{\pi}{2} back into the expression and simplify.

    dxdy=−34csc⁡2(2y−π2)⋅2\frac{dx}{dy} = -\frac{3}{4} \csc^2 \left(2y - \frac{\pi}{2}\right) \cdot 2 dxdy=−3⋅24csc⁡2(2y−π2)\frac{dx}{dy} = -\frac{3 \cdot 2}{4} \csc^2 \left(2y - \frac{\pi}{2}\right) dxdy=−64csc⁡2(2y−π2)\frac{dx}{dy} = -\frac{6}{4} \csc^2 \left(2y - \frac{\pi}{2}\right) dxdy=−32csc⁡2(2y−π2)\frac{dx}{dy} = -\frac{3}{2} \csc^2 \left(2y - \frac{\pi}{2}\right)

    The final answer is −32csc⁡2(2y−π2)-\frac{3}{2} \csc^2 \left(2y - \frac{\pi}{2}\right).

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