Question

http://prntscr.com/kjd9mz

Answer 1

a) it wants the derivative wrt x so you will be taking the derivative of both sides wrt x

d/dx (x) = d/dx (y^2 * sinh(4x) + cosh(y))

the left side is simple, the derivative of x wrt x is simply 1

for the right side, apply the product rule

d/dx (y^2 * sinh(4x) + cosh(y))
=
2y(dy/dx)*sinh(4x) + y^2 * cosh(4x) + sinh(y) (dy/dx)

*** note: for this step, I am using the fact that

the derivative of sinh is cosh and the derivative of cosh is sinh

Answer 2

so from there you have

1 = 2y(dy/dx)*sinh(4x) + y^2 * cosh(4x) + sinh(y) (dy/dx)

simply move the y^2 * cosh(4x) part to the other side and then factor out the dy/dx so you can isolate it and thus provide the solution

Answer 3

also "wrt" means with respect to, sorry if that wasn't clear

Answer 4

b) tanh(x+y)

ngl i had to look this one up but you would do a u-substitution where u = x + y

so it becomes d/du of tanh(u) which, according to the chain rule, becomes

sech^2(u) * d/du * u using the fact that the derivative of tanh is sech^2

plugging u back in gives us

sech^2(x+y) * ∂/∂x (x+y)

∂/∂x of (x+y) is just 1 since the derivative of x is just 1 and the partial derivative of y wrt x is 0

Answer 5

heart eyes Thank you! @Komaeda