Respuesta :
Since y is a function of x, you can basically treat it as another link on the chain rule.
For example, let's say you had sin(y) and wanted to derive it with respect to x. The chain rule states:
[tex] \frac{d}{dx}(sin(y)) = \frac{dy}{dx}* \frac{d}{dy}(sin(y)) = \frac{dy}{dx}*cos(y).[/tex]
So now we can solve the problem.
The left side is a product rule, therefore:
[tex] \frac{d}{dx}(y+cos(\frac{1}{y}))=\frac{d}{dx}(y)*cos(\frac{1}{y})+y*\frac{d}{dx}(cos(\frac{1}{y}) [/tex]
Which, using our method above, is (by the Chain rule) equivalent to:
[tex]\frac{dy}{dx}\frac{d}{dy}(y)*cos(\frac{1}{y})+y*\frac{dy}{dx}\frac{d}{dy}(cos(\frac{1}{y}))[/tex]
And: doing those derivatives with respect to y (and leaving those [tex] \frac{dy}{dx} [/tex]'s alone:
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y*\frac{dy}{dx}(-sin(\frac{1}{y})*\frac{d}{dy}( \frac{1}{y})) [/tex]
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y*\frac{dy}{dx}(-sin(\frac{1}{y})*-y^{-2}) [/tex]
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y^{-1}*\frac{dy}{dx}*sin(\frac{1}{y})[/tex]
And now, for the right side:
[tex] \frac{d}{dx}(8x + 8y) = \frac{d}{dx}8x+\frac{d}{dx}8y = 8 + \frac{dy}{dx}\frac{d}{dy}(8y) = 8 + \frac{dy}{dx}8[/tex]
Simplifying one more time and putting it back the left and right sides into the equation:
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y^{-1}*\frac{dy}{dx}*sin(\frac{1}{y}) = 8 + 8\frac{dy}{dx}[/tex]
For example, let's say you had sin(y) and wanted to derive it with respect to x. The chain rule states:
[tex] \frac{d}{dx}(sin(y)) = \frac{dy}{dx}* \frac{d}{dy}(sin(y)) = \frac{dy}{dx}*cos(y).[/tex]
So now we can solve the problem.
The left side is a product rule, therefore:
[tex] \frac{d}{dx}(y+cos(\frac{1}{y}))=\frac{d}{dx}(y)*cos(\frac{1}{y})+y*\frac{d}{dx}(cos(\frac{1}{y}) [/tex]
Which, using our method above, is (by the Chain rule) equivalent to:
[tex]\frac{dy}{dx}\frac{d}{dy}(y)*cos(\frac{1}{y})+y*\frac{dy}{dx}\frac{d}{dy}(cos(\frac{1}{y}))[/tex]
And: doing those derivatives with respect to y (and leaving those [tex] \frac{dy}{dx} [/tex]'s alone:
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y*\frac{dy}{dx}(-sin(\frac{1}{y})*\frac{d}{dy}( \frac{1}{y})) [/tex]
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y*\frac{dy}{dx}(-sin(\frac{1}{y})*-y^{-2}) [/tex]
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y^{-1}*\frac{dy}{dx}*sin(\frac{1}{y})[/tex]
And now, for the right side:
[tex] \frac{d}{dx}(8x + 8y) = \frac{d}{dx}8x+\frac{d}{dx}8y = 8 + \frac{dy}{dx}\frac{d}{dy}(8y) = 8 + \frac{dy}{dx}8[/tex]
Simplifying one more time and putting it back the left and right sides into the equation:
[tex]\frac{dy}{dx}*cos(\frac{1}{y})+y^{-1}*\frac{dy}{dx}*sin(\frac{1}{y}) = 8 + 8\frac{dy}{dx}[/tex]