Question

Why is d/dx sin(xy) = y cos(xy) and not equal to cos(xy)(xy' + y) ?

Answer 1

There is only xy inside the function.

Answer 2

\(\huge \dfrac{\partial sin(xy) }{\partial x}= y cos(xy) \)

Answer 3

the derivative of xy is only y

Answer 4

Hint
g(h(x)) ' =g'(h(x) h'(x)

Answer 5

To be able to use the chain rule you must have 2 sets of functions with the independent variable

Answer 6

so is it true if I differentiate implicitly on x

x = sin(xy)

I get

1 = ycos(xy)y' ?

Answer 7

you are doing partial differentiation ?

Answer 8

I don't think so. I am trying to do implicit differentiaion on the function

x = sin(xy)

Answer 9

then you need to use product rule!

1 = cos (xy) d/dx (xy)

and for d/dx(xy) use product rule

Answer 10

i see you got
cos(xy)(xy' + y)

and thats correct

Answer 11

d/dx sin(xy) = y cos(xy) <<< incorrect

Answer 12

Okay. I will report to my client that the textbook answer 1/(ycos(xy) = y' is wrong.

Answer 13

lets complete it

x =sin (xy)

x'= cos xy (x y' +yx')

x' =1, as we are differentating w.r.t x

1 =cos xy (x y' +y)

sec xy = xy' +y
y' = (sec (xy) -y) /x

Answer 14

Yes. we got trig variations on the same thing, but in the face of the textbook answer we were unsure.

Answer 15

one reason i can think when we can get
d/dx sin(xy) = y cos(xy)

is when y is given as CONSTANT

if y is function of x, then we are correct

Answer 16

Right, but if the instruction is to differentiate implicitly, then I think it is safe to say y is implicitly defined as function of d

Answer 17

*x