Hi,

Can anyone explain how I can find dy/dx of the following:[

8Ye^(6XY)]=sin(4x)

Many thanks

Bryn

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- Feb 6th 2010, 10:22 AMBrynimplicit differentiation
Hi,

Can anyone explain how I can find dy/dx of the following:[

8Ye^(6XY)]=sin(4x)

Many thanks

Bryn - Feb 6th 2010, 10:40 AMsongoku
Hi Bryn

$\displaystyle 8y~e^{6xy} = \sin(4x)$

$\displaystyle \ln (8y~e^{6xy}) =\ln \sin(4x)$

$\displaystyle \ln(8y)+6xy=\ln \sin(4x)$

Now do the implicit differentiation - Feb 6th 2010, 12:09 PMBryn
I can't seem to proceed easily,

The answer I have worked out is as follows:

(1/8y)(dy/dx)+6y+6x*(dy/dx)=(4cos(4x))/sin(4x)

Using this the question asks me to find the value of dy/dx at the coordinates (0,0).

I found this to be 0 which it seems is wrong, can someone explain to me why this is?

Many thanks