What is the derivative of (x+y+1^2) ?

The derivative of (x+y+1^2) is 2(x+y+1)((dy)/(dx)+1)

What is the first derivative of (x+y+1^2) ?

The first derivative of (x+y+1^2) is 2(x+y+1)((dy)/(dx)+1)

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\frac{d}{d x} ((x + y + 1)^{2})

2 (x + y + 1) (\frac{d y}{d x} + 1)

Solution steps

\frac{d}{d x} ((x + y + 1)^{2})

Apply the chain rule:

2 (x + y + 1) \frac{d}{d x} (x + y + 1)

= 2 (x + y + 1) \frac{d}{d x} (x + y + 1)

\frac{d}{d x} (x + y + 1) = \frac{d y}{d x} + 1

= 2 (x + y + 1) (\frac{d y}{d x} + 1)

f (x) = cos (a^{9} + x^{9})

\int \frac{x}{25 - 4 x ^{2}} d x

\int_{- \infty}^{0} 3^{x} d x

6 y^{^{''}} + 23 y^{^{'}} - 4 y = 0

\frac{d}{d x} ((x - 4) (\frac{x}{2} + 1)^{3})

What is the derivative of (x+y+1^2) ?
The derivative of (x+y+1^2) is 2(x+y+1)((dy)/(dx)+1)
What is the first derivative of (x+y+1^2) ?
The first derivative of (x+y+1^2) is 2(x+y+1)((dy)/(dx)+1)