Question

It is given that m=3x+1 and y= (5/1+m^2).Find dy/dx in terms of x.​

Answer 1

Answer:

${ \underline{ \bf{ \frac{dy}{dx} = - \frac{30m}{ {(1 + {m}^{2}) }^{2} } }}}$

Step-by-step explanation:

${ \bf{m = 3x + 1}} \\ { \sf{ \frac{dm}{dx} = 3 }} \\ \\ { \bf{y = \frac{5}{1 + {m}^{2} } }} \\ \\ { \tt{ \frac{dy}{dm} = \frac{ - 10m}{ {(1 + {m}^{2} )}^{2} } }}$

Using chain rule:

${ \boxed{ \bf{ \frac{dy}{dx} = \frac{dy}{dm}. \frac{dm}{dx} }}}$

${ \sf{ \frac{dy}{dx} = - \frac{10m}{ {(1 + {m}^{2}) }^{2} } \times 3}} \\ \\ { \sf{ \frac{dy}{dx} = - \frac{30m}{ {(1 + {m}^{2}) }^{2} } }}$

${ \underline{ \sf{ \blue{christ \:† \: alone }}}}$

Answer 2

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:

$y=\frac{5}{1+(3x+1)^2}$ which simplifies to

$y=\frac{5}{1+9x^2+6x+1}$ and a bit more to

$y=\frac{5}{9x^2+6x+2}$ and now we're ready to find the derivative. Using the quotient rule:

$y'=\frac{(9x^2+6x+2)(0)-[5(18x+6)]}{(9x^2+6x+2)^2}$ which simplifies to

$y'=\frac{-90x-30}{(9x^2+6x+2)^2}$ or, equally:

$y'=-\frac{90x+30}{(9x^2+6x+2)^2}$