What is the inverse of D/(Dx)(x^2+4x+1) ?

The inverse of D/(Dx)(x^2+4x+1) is (-4+x+sqrt(x^2-8x+12))/2 ,(-4+x-sqrt(x^2-8x+12))/2

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inverse of D/(Dx)(x^2+4x+1)

inverse

\frac{D}{D x} (x^{2} + 4 x + 1)

\frac{- 4 + x + x ^{2} - 8 x + 12}{2}, \frac{- 4 + x - x ^{2} - 8 x + 12}{2}

Solution steps

y = \frac{D}{D x} (x^{2} + 4 x + 1)

Replace

x

with

y

x = \frac{D}{Dy} (y^{2} + 4 y + 1)

Solve for

y, x = \frac{D}{Dy} (y^{2} + 4 y + 1)

\frac{- 4 + x + x ^{2} - 8 x + 12}{2}, \frac{- 4 + x - x ^{2} - 8 x + 12}{2}

f (x) = lo g_{10} (x - 2) - 7

7 sin (x) - 1

sin (4 5^{\circ})

f (x) = \frac{( 2 x + 4 )}{( x - 1 )}

f (x) = - \frac{4 x - 3}{x}

What is the inverse of D/(Dx)(x^2+4x+1) ?
The inverse of D/(Dx)(x^2+4x+1) is (-4+x+sqrt(x^2-8x+12))/2 ,(-4+x-sqrt(x^2-8x+12))/2