`4cos(x)sin(y) = 1` Find `dy/dx` by implicit differentiation.

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hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

`4cos(x)*sin(y) = 1`

``Differentiating both sides w.r.t 'x' we get

`-4sin(x)sin(y) + 4cos(x)cos(y)(dy/dx) = 0`

`or, dy/dx = [sin(x)sin(y)]/[cos(x)cos(y)]`

`or, dy/dx = y' = tan(x)*tan(y)`

``

` `

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kandukurimaths | Student, Graduate | (Level 1) Honors

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4cos(x)sin(y)=1

sin(y)=(1/4)sec(x)

apply derivative both side with respect x

cos(y)(dy/dx)=(1/4)sec(x)tan(x)

dy/dx=(1/4)sec(x)tan(x)/cos(y)

dy/dx=(sin(y)tan(x)/cos(y))

dy/dx=(tan(y)tan(x))