Answer to Question #103971 in Calculus for Martin

Question #103971
Find d²y/dx² for y=csc(x)
1
Expert's answer
2020-03-06T16:48:24-0500

Solution:

1.      This expression can be written as:

d2yd2x(csc(x))=dydx(dydxcsc(x))\frac{d^2y}{d^2x}(\csc(x)) = \frac{dy}{dx}(\frac{dy}{dx}\csc(x))


2.      Find the first derivative of the specified function:

dydxcsc(x)=dydx(1sin(x))=(1)×sin(x)1×(sin(x))sin2(x)=cos(x)sin2(x)\frac{dy}{dx}\csc(x) = \frac{dy}{dx}(\frac{1}{\sin(x)}) = \frac{(1^{\prime})×\sin(x) - 1×(\sin(x))^{\prime}}{\sin^{2}(x)} = - \frac{\cos(x)}{\sin^{2}(x)}

3.      Next, find the derivative of the resulting function:

dydx(cos(x)sin2(x))=(cos(x))×sin2(x)(cos(x))×(sin2(x))(sin2(x))2=\frac{dy}{dx}(\frac{-\cos(x)}{\sin^{2}(x)}) = \frac{(-\cos(x))^{\prime}×\sin^{2}(x) - (-\cos(x))×(\sin^{2}(x))^{\prime}}{(\sin^{2}(x))^{2}} =

=sin(x)×sin2(x)+cos(x)×(2sin(x)cos(x))sin4(x)=sin(x)×(sin2(x)+2cos2(x))sin4(x)== \frac{\sin(x) × \sin^{2}(x) + \cos(x) × (2 \sin(x)\cos(x))}{\sin^{4}(x)} = \frac{\sin(x)×(\sin^{2}(x) + 2\cos^{2}(x))}{\sin^{4}(x)} =

=sin2(x)+cos2(x)+cos2(x)sin3(x)=1+cos2(x)sin3(x)=1sin3(x)(1+cos2(x))== \frac{\sin^{2}(x) + \cos^{2}(x) +\cos^{2}(x) }{\sin^{3}(x)} = \frac{1+ \cos^{2}(x)}{\sin^{3}(x)} = \frac{1}{\sin^{3}(x)}(1 + \cos^{2}(x)) =

=csc3(x)(1+cos2(x))= \csc^{3}(x)(1+\cos^{2}(x))


Answer:

 d2yd2x(csc(x))=(cos2(x)+1)csc3(x)\frac{d^2y}{d^2x}(\csc(x)) = (\cos^{2}(x) + 1) \csc^{3}(x)


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Comments

Assignment Expert
09.03.20, 16:13

Dear Ogumo, please use the panel for submitting new questions.

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09.03.20, 15:38

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Ogumo
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