# We are given that y = t^2 , t = 3p+5 and p = 6x^2 . What is dy/dx?

william1941 | College Teacher | (Level 3) Valedictorian

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Here we use the chain rule. So we express dy/dx as differentials in terms of t, p and x

dy/dx = (dy/ dt) (dt/ dp) (dp/ dx)

Now y = t^2 => dy /dt = 2t

t = 3p + 5 => dt /dp = 3

p = 6x^2 => dp / dx = 12x

Using these dy/dx = (dy/ dt) (dt/ dp) (dp/ dx)

=> 2t * 3 * 12x

substitute t = 3p+5

=> 2*(3p+5)* 36x

=> (6p+ 10)*36x

substitute p = 6 x^2

=> (36x^2 +10)*36x

=> 1296 x^3 +360 x

Therefore the required result is 1296 x^3 +360 x

neela | High School Teacher | (Level 3) Valedictorian

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y =t^2 .

t =3p+5

p = 6x^2

are the give relations. To find dy/dx.

y  = t^2.

y  = (3p+5)^2 , as t = 3p+5

y = {3(6x^2)+5}^2 = (18x^2+5)^2

y' = 2(18x^2+5)^(2-1) *(18x^2+5)'

y' = 2(18x^2+5)(18*2x)

y' = 72(18x^2+5)x.

Alternative method:

Therefore , y = (18x^2+5)^2 = 2

To solve this we try to find y  in terms of p and x interms p and then find dy/dp  and dx/dp. Then we use :

dy/dx = (dy/dp)/(dx/dp).

y = t^2 and t = 3p+5. We eliminate t and find y in terms of p.

y= (3p+5)^2.

We differentiate  y wrt p:

y' ={(3p+5)^2}'

y' = 2(3p+5)^(2-1)*(3p+5)'

y' =2(3p+5)*3

y' = 6(3p+5)............(1).

6x^2 = p

We differentiate both sides of the equation  wrt p  :

6*2x dx/dp = 1

dx/dp = 1/12x............(2).

Therefore using(1) and(2):

dy/dx = (dy/dp)/(dx/dp) =  6(3p+5)/(1/12x) = 12*6(3p+5)

Therefore

dy/dx = 72(3p+5)x

dy/dx = 72(18x^2+5)x as, p = 6x^2.