Determine derivative of y=(x+3)^2+2(x+3)(x-4)+(x-4)^2

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We have to find the derivative of y = (x+3)^2 + 2(x+3)(x-4) + (x-4)^2

y = (x+3)^2 + 2(x+3)(x-4) + (x-4)^2

Use the relation [x^n]' = n*x^(n - 1) and the chain rule

y' = 2( x + 3) + 2*( x + 3) + 2* ( x - 4) + 2*(x - 4)

=> 2x + 6 + 2x + 6 + 2x - 8 + 2x - 8

=> 8x - 4

The required derivative is 8x - 4

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giorgiana1976 | Student

If we'll note (x+3) by a and (x-4) by b, and we'll re-write the equation, we'll get a perfect square:

y=a^2 + 2ab + b^2

y = (a+b)^2

y = (x+3+x-4)^2

We'll combine like terms:

y = (2x-1)^2

Now, we'll differentiate both sides, with respect to x:

dy/dx = 2(2x-1)*(2x-1)'

dy/dx = 2(2x-1)*2

dy/dx = 4(2x-1)

W'll remove the brackets:

dy/dx = 8x - 4

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