hi I am just starting to learn deferentials and I am solving this problem.

[feral wedge]

$$\dv{\qty(e^u)}{u} = e^u$$
$$\dv{\qty(e^u)}{x} = \dv{\qty(e^u)}{u} \cdot \dv{u}{x}$$

[prisma idol]

King Leo

[void blade]

but then, I am confused. I am just doing d(e^u) = e^u * du

if that makes sense

So there is deferentiation with respect to a variable and then implicit deferentiation, right?

and for the second row I get du with respect to x and y

not just x

[feral wedge]

Hold on, im having trouble understanding what exactly youre asking

[void blade]

hmm, me too a little bit

well

lets just look at right side e^(-y *cos x )

[feral wedge]

$$e^{-y \cos(x)}$$

[prisma idol]

King Leo

[void blade]

this a function F(X,Y) = e^(-y*cosx)

so, when we deferentiate with respect to x we get

but I dont understand what is happening in the implicit deferentiation

[feral wedge]

You need to use the chain rule, assuming y is a function of x

[void blade]

okay

so it wouldnt be fair to just make a function f(x,y)

okay

hmm I don't know what I am asking then really

but I still don't understand what I am doing right now

feels more just like doing what I am told rather than understanding

so when I do implicit deferentiation of both side of

what exactly is happening

[feral wedge]

So do you want me to explain the implicit differentiation of this

[void blade]

more like how it works

like for example if i were to explain something simpler

x - 3 < 2

x < 5

for this example we add 3 to both sides

rather than doing something random and arbitrary

I dont understand how the deferentiation of both sides works

[feral wedge]

$$e^{-y \cos(x)} \cdot \dv{(-y \cos(x))}{x}$$
$$e^{-y \cos(x)} \cdot \qty(\cos(x)\dv{-y}{x} - y \dv{\cos(x)}{x})$$

[void blade]

implicitly

okay

why is when we deferentiat sinx implicitly we get

cosx * dx

[prisma idol]

King Leo

[feral wedge]

I gtg sry

[void blade]

its okay

thanks for trying to help

okay never mind i got it

/done