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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:
[tex]\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)}[/tex]
Apply the product rule to differentiate that term at the left-hand side:
[tex]\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}[/tex]
Now, multiply out the terms to get rid of the brackets at the left-hand
dy
side, and then isolate —— :
dx
[tex]\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}[/tex]
and there it is.
I hope this helps. =)
Tags: implicit function derivative implicit differentiation chain product rule differential integral calculus
