# How to find dy/dx from × = ln cos (xy)?

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#### keni4eva

##### Newbie level 5 Can someone pls help me solve this problem.

× = ln cos (xy)

find dy/dx

Thanks

#### piash

##### Member level 1 Re: PLS HELP

implicit differentiation

x = ln {cos(xy)}
or, dx/dx = (1/ cos(xy)). {d/dx (cos(xy))}
or, 1 = (1/ cos(xy)). (-sin(xy)). {d/dx(xy)}
or, 1 = (-tan(xy)).1.(dy/dx)
or, dy/dx = -cot(xy)

#### salam2000

##### Member level 3 Re: PLS HELP

{d/dx(xy)}=(dy/dx)

???????????
d/dx(xy)=y+xdy/dx
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x = ln {cos(xy)}
or, dx/dx = (1/ cos(xy)). {d/dx (cos(xy))}
or, 1 = (1/ cos(xy)). (-sin(xy)). {d/dx(xy)}
or, 1 = (-tan(xy)).1.(y+x dy/dx)
-cot(xy)=y+x dy/dx
or, dy/dx = -(cot(xy)+y)/x

#### piash

##### Member level 1 Re: PLS HELP

thankz salam for ur correction, i forgot to do that

#### keni4eva

##### Newbie level 5 Re: PLS HELP

Let me know what the correction is

#### Old Nick Re: PLS HELP

keni4eva said:
Let me know what the correction is

That'll be it written a couple of posts up.

#### Dmitrij Re: PLS HELP

The initial function is given in inevident form - which means, that there is no any expression, which shows the real dependence of y from x. Therefore to find the 1-st derivative you should differentiate both parts by x and lead out the corresponding result:

1 = [1/cos(xy)] * -sin(xy) (y+y'*x);

1 = -tg(xy)*(y+y'*x)

y' = [-ctg(xy)-y]/x

With respect,

Dmitrij

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