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Answer:
[tex]{ \underline{ \bf{ \frac{dy}{dx} = - \frac{30m}{ {(1 + {m}^{2}) }^{2} } }}}[/tex]
Step-by-step explanation:
[tex]{ \bf{m = 3x + 1}} \\ { \sf{ \frac{dm}{dx} = 3 }} \\ \\ { \bf{y = \frac{5}{1 + {m}^{2} } }} \\ \\ { \tt{ \frac{dy}{dm} = \frac{ - 10m}{ {(1 + {m}^{2} )}^{2} } }}[/tex]
Using chain rule:
[tex]{ \boxed{ \bf{ \frac{dy}{dx} = \frac{dy}{dm}. \frac{dm}{dx} }}}[/tex]
[tex]{ \sf{ \frac{dy}{dx} = - \frac{10m}{ {(1 + {m}^{2}) }^{2} } \times 3}} \\ \\ { \sf{ \frac{dy}{dx} = - \frac{30m}{ {(1 + {m}^{2}) }^{2} } }}[/tex]
[tex]{ \underline{ \sf{ \blue{christ \:† \: alone }}}}[/tex]
Answer:
Step-by-step explanation:
Never saw a problem presented in this way in all my years of teaching calculus. But I'm thinking that we need to sub that given expression for m into the equation for y and get everything into y in terms of x in order to find the derivative. I see no other way that makes sense. Can't find the derivative of y in terms of x if there's an m in there. Making that substitution:
[tex]y=\frac{5}{1+(3x+1)^2}[/tex] which simplifies to
[tex]y=\frac{5}{1+9x^2+6x+1}[/tex] and a bit more to
[tex]y=\frac{5}{9x^2+6x+2}[/tex] and now we're ready to find the derivative. Using the quotient rule:
[tex]y'=\frac{(9x^2+6x+2)(0)-[5(18x+6)]}{(9x^2+6x+2)^2}[/tex] which simplifies to
[tex]y'=\frac{-90x-30}{(9x^2+6x+2)^2}[/tex] or, equally:
[tex]y'=-\frac{90x+30}{(9x^2+6x+2)^2}[/tex]