What is the derivative of-(x+p^2+a) ?

The derivative of-(x+p^2+a) is -2x-2p(dp)/(dx)-2(x(dp)/(dx)+p)

What is the first derivative of-(x+p^2+a) ?

The first derivative of-(x+p^2+a) is -2x-2p(dp)/(dx)-2(x(dp)/(dx)+p)

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\frac{d}{d x} (- (x + p)^{2} + a)

- 2 x - 2 p \frac{d p}{d x} - 2 (x \frac{d p}{d x} + p)

Solution steps

\frac{d}{d x} (- (x + p)^{2} + a)

Simplify

- (x + p)^{2} + a : - x^{2} - 2 x p - p^{2} + a

= \frac{d}{d x} (- x^{2} - 2 x p - p^{2} + a)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d x} (x^{2}) - \frac{d}{d x} (2 x p) - \frac{d}{d x} (p^{2}) + \frac{d a}{d x}

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (2 x p) = 2 (p + x \frac{d p}{d x})

\frac{d}{d x} (p^{2}) = 2 p \frac{d p}{d x}

\frac{d a}{d x} = 0

= - 2 x - 2 (p + x \frac{d p}{d x}) - 2 p \frac{d p}{d x} + 0

Simplify

= - 2 x - 2 p \frac{d p}{d x} - 2 (x \frac{d p}{d x} + p)

\frac{1}{( 1 - t ^{2} )}

5 e^{3} x

arcsin (3^{\frac{1}{2}}) x

(x^{3} + 4 x)^{3}

9 cos^{3} (x)

What is the derivative of-(x+p^2+a) ?
The derivative of-(x+p^2+a) is -2x-2p(dp)/(dx)-2(x(dp)/(dx)+p)
What is the first derivative of-(x+p^2+a) ?
The first derivative of-(x+p^2+a) is -2x-2p(dp)/(dx)-2(x(dp)/(dx)+p)