How do you implicitly differentiate $csc (\frac{x^{2}}{y^{2}}) = e^{x y}$ $csc(x^2/y^2)=e^(xy)$ ?

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How do you implicitly differentiate $csc (\frac{x^{2}}{y^{2}}) = e^{x y}$ $csc(x^2/y^2)=e^(xy)$ ?

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Answer 1 · 2017-09-02T15:45:55

$\frac{d y}{d x} = \frac{2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) + y^{3} e^{x y}}{x y (2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) - y e^{x y})}$ $dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))$

Explanation:

Implicit differentiation is no different from explicit differentiation. Just remember that differentiating a function of $y$ $y$ causes the chain rule to be in effect. Recall also that $\frac{d}{d x} csc x = - csc x cot x$ $d/dxcscx=-cscxcotx$ and $\frac{d}{d x} e^{x} = e^{x}$ $d/dxe^x=e^x$ .

$csc (\frac{x^{2}}{y^{2}}) = e^{x y}$ $csc(x^2/y^2)=e^(xy)$

$\frac{d}{d x} csc (\frac{x^{2}}{y^{2}}) = \frac{d}{d x} e^{x y}$ $d/dxcsc(x^2/y^2)=d/dxe^(xy)$

$- csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) \cdot \frac{d}{d x} (x^{2} y^{- 2}) = e^{x y} \cdot \frac{d}{d x} (x y)$ $-csc(x^2/y^2)cot(x^2/y^2)*d/dx(x^2y^-2)=e^(xy)*d/dx(xy)$

Use the product rule to find these derivatives. Recall that while $\frac{d}{d x} x^{2} = 2 x$ $d/dxx^2=2x$ , $\frac{d}{d x} y^{2} = 2 y \frac{d y}{d x}$ $d/dxy^2=2ydy/dx$ .

$- csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) (2 x y^{- 2} - 2 x^{2} y^{- 1} \frac{d y}{d x}) = e^{x y} (y + x \frac{d y}{d x})$ $-csc(x^2/y^2)cot(x^2/y^2)(2xy^-2-2x^2y^-1dy/dx)=e^(xy)(y+xdy/dx)$

Expanding and rearranging to group $\frac{d y}{d x}$ $dy/dx$ terms:

$⎛ ⎜ ⎝ \frac{2 x^{2} csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}})}{y} - x e^{x y} ⎞ ⎟ ⎠ \frac{d y}{d x} = \frac{2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}})}{y^{2}} + y e^{x y}$ $((2x^2csc(x^2/y^2)cot(x^2/y^2))/y-xe^(xy))dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2))/y^2+ye^(xy)$

Common denominators:

$⎛ ⎜ ⎝ \frac{2 x^{2} csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) - x y e^{x y}}{y} ⎞ ⎟ ⎠ \frac{d y}{d x} = \frac{2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) + y^{3} e^{x y}}{y^{2}}$ $((2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))/y)dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2$

Solving for $\frac{d y}{d x}$ $dy/dx$ :

$\frac{d y}{d x} = \frac{2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) + y^{3} e^{x y}}{y^{2}} \cdot \frac{y}{2 x^{2} csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) - x y e^{x y}}$ $dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/y^2*y/(2x^2csc(x^2/y^2)cot(x^2/y^2)-xye^(xy))$

$\frac{d y}{d x} = \frac{2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) + y^{3} e^{x y}}{x y (2 x csc (\frac{x^{2}}{y^{2}}) cot (\frac{x^{2}}{y^{2}}) - y e^{x y})}$ $dy/dx=(2xcsc(x^2/y^2)cot(x^2/y^2)+y^3e^(xy))/(xy(2xcsc(x^2/y^2)cot(x^2/y^2)-ye^(xy))$

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How do you implicitly differentiate $csc (\frac{x^{2}}{y^{2}}) = e^{x y}$ $csc(x^2/y^2)=e^(xy)$ ?

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How do you implicitly differentiate $csc (\frac{x^{2}}{y^{2}}) = e^{x y}$ $csc(x^2/y^2)=e^(xy)$ ?