Answer to Question #270981 in Calculus for Kamo

Question

Consider the interesting curve below

which is described by the equation

cosh (sinh-¹(v²))=√1+x² dy Determine an expression for y'= (your expression will contain both x and y dx functions). Use your calculations to answer questions 1 to 4 below.

1. The derivative with regards to x of sinh ¹ (y² is

2. Using the chain rule the derivative of cosh (sinh-¹(y²)) with regards to x is

3. The derivative of √√1+x² is

dy

4. The simplified version of y'= in terms of x and y is

dx

2

3

Answer 1

$$\cosh(\sinh^{-1}(y^2))=\sqrt{1+x^2}$$

1.

$$(\sinh^{-1}(y^2))'_x=\dfrac{1}{\sqrt{1+y^4}}(2y)y'$$

2.

$$(\cosh(\sinh^{-1}(y^2)))'_x$$

$$=\sinh(\sinh^{-1}(y^2))\cdot(\sinh^{-1}(y^2))'_x$$

$$=y^2(\dfrac{1}{\sqrt{1+y^4}}(2y)y')=\dfrac{2y^3}{\sqrt{1+y^4}}y'$$

3.

$$(\sqrt{1+x^2})'_x=\dfrac{2x}{2\sqrt{1+x^2}}=\dfrac{x}{\sqrt{1+x^2}}$$

4.

$$\dfrac{2y^3}{\sqrt{1+y^4}}y'=\dfrac{x}{\sqrt{1+x^2}}$$

$$y'=\dfrac{x\sqrt{1+y^4}}{2y^3\sqrt{1+x^2}}$$