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Let D(f(x)) = f'(x) be the derivative of f(x).
We take the derivative of both sides of the equation:
`D(sin(x^4y^4)) = D(x)`
`cos(x^4y^4) D(x^4y^4) = 1`
`cos(x^4y^4) [4x^4y^3 (dy)/(dx) + 4y^4x^3] = 1`
`(dy)/(dx) = [(1/(cos(x^4y^4))) - 4y^4x^3]/(4x^4y^3)`
Here, we used the chain rule in evaluating the derivative of `sin(x^4y^4)`.
We can think of sin(x) as the outer function, with inner function `x^4y^4` . The chain rule states that the derivative of f(g(x)) with respect to x is the derivative of f(x) multiplied by the derivative of g(x).
In this case, we first got the derivative of the outer function:
`D(sin(x^4y^4)) = cos(x^4y^4)`
and multipled it with the derivative of the inner function (which we got by applying the product rule, and noting that we are getting the implicit derivative.
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