Use-implicit-differentiation-to-find-d-2-y-dx-2-for-siny-x-

Question Number 184040 by pete last updated on 02/Jan/23

Use implicit differentiation to find \frac{d^{2} y}{{dx}^{2}}

for \sin y = x

Answered by cortano2 last updated on 02/Jan/23

y^{'} \cos y = 1

y^{″} (- \sin y) = 0

y^{″} = 0

Answered by Yhusuph last updated on 02/Jan/23

\frac{d y}{d x} c o s y = 1

\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}

\frac{d^{2} y}{{d x}^{2}} = \frac{- (- 2 {(1 - x^{2})}^{- \frac{1}{2}})}{1 - x^{2}}

\frac{d^{2} y}{{d x}^{2}} = \frac{2}{{(1 - x^{2})}^{\frac{3}{2}}}

\frac{d^{2} y}{{d x}^{2}} = \frac{2}{{(^{2} \sqrt{1 - x^{2}})}^{3}}

\partial e n k e n L a s t β o r n

M e n t o r : P r o f f y e m p h y

Commented by Frix last updated on 02/Jan/23

This cannot be true .

y = \arcsin x \Rightarrow \frac{d^{2} y}{{d x}^{2}} = \frac{x}{{(1 - x^{2})}^{\frac{3}{2}}}

Answered by mr W last updated on 03/Jan/23

\sin y = x

y^{'} \cos y = 1

y^{″} \cos y - {(y^{'})}^{2} \sin y = 0

y^{″} = {(y^{'})}^{2} \frac{\sin y}{\cos y} = \frac{\tan y}{\cos^{2} y} = \frac{\sin y}{\cos^{3} y} = \frac{\sin y}{{(1 - \sin^{2} y)}^{\frac{3}{2}}}

y^{″} = \frac{x}{{(1 - x^{2})}^{\frac{3}{2}}}

Commented by pete last updated on 02/Jan/23

Thank you Mr . W, but the answer given

is : \frac{d^{2} y}{{dx}^{2}} = \sec^{2} ytany

Commented by cortano1 last updated on 03/Jan/23

f r o m \sin y = x w e g e t {\begin{cases} \sec y = \frac{1}{\sqrt{1 - x^{2}}} \\ \tan y = \frac{x}{\sqrt{1 - x^{2}}} \end{cases}

n o w y^{″} = \frac{x}{(1 - x^{2}) \sqrt{1 - x^{2}}}

= {[\frac{1}{\sqrt{1 - x^{2}}}]}^{2} . \frac{x}{\sqrt{1 - x^{2}}} = \sec^{2} y \tan y

Commented by mr W last updated on 03/Jan/23

y o u c a n e x p r e s s y^{″} i n t e r m s o f y o r

i n t e r m s o f x . y o u s e e b o t h a b o v e .

y^{″} = {(y^{'})}^{2} \frac{\sin y}{\cos y} = \frac{\tan y}{\cos^{2} y} = \frac{\sin y}{\cos^{3} y} = \frac{\sin y}{{(1 - \sin^{2} y)}^{\frac{3}{2}}}

y^{″} = \frac{x}{{(1 - x^{2})}^{\frac{3}{2}}}

Commented by pete last updated on 04/Jan/23

Thanks very much sir

Answered by manxsol last updated on 02/Jan/23

Commented by mr W last updated on 03/Jan/23

m a y b e n o t a l l p e o p l e c a n r e a d y o u r

s o l u t i o n p r o p e r l y . t h i s i s h o w y o u r

p o s t l o o k s l i k e :

Commented by mr W last updated on 03/Jan/23

Commented by ARUNG_Brandon_MBU last updated on 03/Jan/23