please find `dy/dx`  using implicit differentiation, and please explain the use of the chain rule as a method used to obtain the answer. `sin(x^4y^4)=x`



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Posted on (Answer #1)

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)` 

We have:

`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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