Question

If cos(xy) = 3x+1 , find dy/dx

Answer 1

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          dy
Find  ——  for an implicit function:
          dx

cos(xy) = 3x + 1.


First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

$\mathsf{\dfrac{d}{dx}\big[cos(xy)\big]=\dfrac{d}{dx}(3x+1)}\\\\\\ \mathsf{-\,sin(xy)\cdot \dfrac{d}{dx}(xy)=\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(1)}$


Apply the product rule to differentiate that term at the left-hand side:

$\mathsf{-\,sin(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=3+0}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=3}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[y+x\cdot \dfrac{dy}{dx}\right]=3}$

   

Now, multiply out the terms to get rid of the brackets at the left-hand
                                       dy
side, and then isolate  —— :
                                       dx

$\mathsf{-\,sin(xy)\cdot y-sin(xy)\cdot x\cdot \dfrac{dy}{dx}=3}\\\\\\ \mathsf{-\,y\,sin(xy)-x\,sin(xy)\cdot \dfrac{dy}{dx}=3}\\\\\\ \mathsf{-\;x\,sin(xy)\cdot \dfrac{dy}{dx}=3+y\,sin(xy)}\\\\\\\\ \therefore~~\mathsf{\dfrac{dy}{dx}=\dfrac{3+y\,sin(xy)}{-\;x\,sin(xy)}\qquad\quad for~~x\,sin(xy)\ne 0\qquad\quad\checkmark}$


and there it is.


I hope this helps. =)


Tags:  implicit function derivative implicit differentiation chain product rule differential integral calculus