Question

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

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=(5)/(1+(3x+1)^2)` which simplifies to

`y=(5)/(1+9x^2+6x+1)` and a bit more to

`y=(5)/(9x^2+6x+2)` and now we're ready to find the derivative. Using the quotient rule:

`y'=((9x^2+6x+2)(0)-[5(18x+6)])/((9x^2+6x+2)^2)` which simplifies to

`y'=(-90x-30)/((9x^2+6x+2)^2)` or, equally:

`y'=-(90x+30)/((9x^2+6x+2)^2)`

Answer 2

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

Explanation:

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

Using chain rule:

`{ \boxed{ \bf{ (dy)/(dx) = (dy)/(dm). (dm)/(dx) }}}`

`{ \sf{ (dy)/(dx) = - \frac{10m}{ {(1 + {m}^(2)) }^(2) } * 3}} \\ \\ { \sf{ (dy)/(dx) = - \frac{30m}{ {(1 + {m}^(2)) }^(2) } }}`

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