Y=3x^2+6x-7/(x+1)^2 Show that dy/dx = 20/(x+1)^3

1 answer:

3 0

Apply the quotient rule
d(uv) = v du/dx - u dv/dx
dx ----------------------
v^2

du/dx = 6x + 6
dv/dx = 2(x + 1)
v^2 = (x + 1)^4 so we have

dy dx = (x + 1)^2 .* (6x + 6) - (3x^2 + 6x - 7) * 2(x + 1) / (x + 1)^4

= (x + 1) [(6(x + 1)^2 - 2(3x^2 + 6x - 7) ] / (x + 1)^4

= (x^2 + 12x +6 - 6x^2 - 12x + 14 ) / (x + 1)^3

= 20 / (x + 1)^3 Answer

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Solve for x:a(a²+b²)x²+b²x-a

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x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

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Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$