Question

dy/dx+tanx tany=cosx secy

Answer 1

multiply by cosy and you have

cosy y' + tanx siny = cosx
Now let u = siny and you have
u' + tanx u = cosx
That should look more familiar.

Answer 2

Yes

Answer 3

Great 👌

Answer 4

[AI-generated answer — Clown Bot]

Well, well, well, someone dropped a calculus bomb on me! Let's see if we can diffuse it with some comedic relief.

Alright, so you want to find dy/dx. Let's break it down step by step, like a clown trying to juggle chainsaws.

First, we have dy/dx + tan(x)tan(y) = cos(x)sec(y). But how do we separate dy/dx? It's like trying to untangle your headphones after leaving them alone for five seconds!

We can start by moving the tan(x)tan(y) to the other side. Now, we have dy/dx = cos(x)sec(y) - tan(x)tan(y). It's like trying to solve a puzzle with missing pieces, but we'll piece it together!

Now, let's simplify it further. Remember that cos(x)sec(y) can be rewritten as cos(x)/cos(y) because sec(y) is equal to 1/cos(y). Alright, let's substitute! We get dy/dx = cos(x)/cos(y) - tan(x)tan(y).

Alright, now it's time to put on our clown nose and perform some trigonometric acrobatics. We know that tan(x) = sin(x)/cos(x) and tan(y) = sin(y)/cos(y). Let's substitute these in! We get dy/dx = cos(x)/cos(y) - (sin(x)/cos(x))(sin(y)/cos(y)).

Now, let's simplify it even further, just like a clown squeezing into a tiny car. We can cross-multiply and combine the terms: dy/dx = (cos(x)cos(x) - sin(x)sin(y))/(cos(x)cos(y)).

Hey, what's this? Cos(x)cos(x) - sin(x)sin(y)? That's just the good old cosine of the sum! So, we can simplify it as dy/dx = cos(x + y)/(cos(x)cos(y)). Hooray! We've defused the calculus bomb!

Now, I hope my clownish approach didn't make you dizzy! Just remember to keep a sense of humor while tackling these math problems. Good luck with the rest of your calculations!

Answer 5

[AI-generated answer — Explain Bot]

To find the derivative of the given expression, we can use the product rule and chain rule.

Let's break down the given expression:

dy/dx + tan(x) * tan(y) = cos(x) * sec(y)

Now, let's differentiate both sides with respect to x:

d/dx(dy/dx) + d/dx(tan(x) * tan(y)) = d/dx(cos(x) * sec(y))

To simplify, we need to use the chain rule. Remember that the derivative of tan(x) is sec^2(x) and the derivative of tan(y) is sec^2(y) * dy/dx.

Using the product rule, the left-hand side differentiation becomes:

d^2y/dx^2 + (sec^2(x) * tan(y) + tan(x) * sec^2(y) * dy/dx)

The right-hand side differentiation becomes:

-d/dx(sin(x)) * sec(y) - cos(x) * sec(y) * tan(y) * dy/dx

Now, let's isolate d^2y/dx^2 by moving all other terms to the right-hand side:

d^2y/dx^2 - (sec^2(x) * tan(y) + tan(x) * sec^2(y) * dy/dx) = -d/dx(sin(x)) * sec(y) - cos(x) * sec(y) * tan(y) * dy/dx

Finally, we can factor out dy/dx from the left-hand side and divide both sides by (sec^2(x) * tan(y) + tan(x) * sec^2(y)):

d^2y/dx^2 - (sec^2(x) * tan(y) + tan(x) * sec^2(y) * dy/dx) / (sec^2(x) * tan(y) + tan(x) * sec^2(y)) = (-d/dx(sin(x)) * sec(y) - cos(x) * sec(y) * tan(y) * dy/dx) / (sec^2(x) * tan(y) + tan(x) * sec^2(y))

Simplifying further:

d^2y/dx^2 - dy/dx = -d/dx(sin(x)) / (sec(x) * tan(y)) - cos(x) * tan(y) / (sec(y) * sec(x) * tan(y))

Now, we can substitute the values of sin(x), cos(x), sec(x), and sec(y) to simplify the equation further. After simplification, we will have the second derivative of y with respect to x, d^2y/dx^2.