Answer to Question #270947 in Calculus for pakie

Question #270947

Consider the interesting curve below which is described by the equation ( ( )) 12 2 cosh sinh y 1 x − = + Determine an expression for dy y' dx = (your expression will contain both x and y functions). Use your calculations to answer questions 1 to 4 below. 1. The derivative with regards to x of ( ) 1 2 sinh y − is 2 2. Using the chain rule the derivative of ( ( )) 1 2 cosh sinh y − with regards to x is 3 3. The derivative of 2 1 x + is 2 4. The simplified version of dy y' dx = in terms of x and y is


1
Expert's answer
2021-11-25T07:27:36-0500
cosh(sinh1(y2))=1+x2\cosh(\sinh^{-1}(y^2))=\sqrt{1+x^2}

1.


(sinh1(y2))x=11+y4(2y)y(\sinh^{-1}(y^2))'_x=\dfrac{1}{\sqrt{1+y^4}}(2y)y'

2.


(cosh(sinh1(y2)))x(\cosh(\sinh^{-1}(y^2)))'_x

=sinh(sinh1(y2))(sinh1(y2))x=\sinh(\sinh^{-1}(y^2))\cdot(\sinh^{-1}(y^2))'_x

=y2(11+y4(2y)y)=2y31+y4y=y^2(\dfrac{1}{\sqrt{1+y^4}}(2y)y')=\dfrac{2y^3}{\sqrt{1+y^4}}y'

3.


(1+x2)x=2x21+x2=x1+x2(\sqrt{1+x^2})'_x=\dfrac{2x}{2\sqrt{1+x^2}}=\dfrac{x}{\sqrt{1+x^2}}

4.


2y31+y4y=x1+x2\dfrac{2y^3}{\sqrt{1+y^4}}y'=\dfrac{x}{\sqrt{1+x^2}}

y=x1+y42y31+x2y'=\dfrac{x\sqrt{1+y^4}}{2y^3\sqrt{1+x^2}}


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