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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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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