What is the d/(da)(a^5+a^2b^3-a^3b^2-b^5) ?

The d/(da)(a^5+a^2b^3-a^3b^2-b^5) is 5a^4-3a^2b^2+2ab^3

What is the first d/(da)(a^5+a^2b^3-a^3b^2-b^5) ?

The first d/(da)(a^5+a^2b^3-a^3b^2-b^5) is 5a^4-3a^2b^2+2ab^3

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\frac{d}{d a} (a^{5} + a^{2} b^{3} - a^{3} b^{2} - b^{5})

5 a^{4} - 3 a^{2} b^{2} + 2 a b^{3}

Solution steps

\frac{d}{d a} (a^{5} + a^{2} b^{3} - a^{3} b^{2} - b^{5})

Apply the Sum/Difference Rule:

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

= \frac{d}{d a} (a^{5}) + \frac{d}{d a} (a^{2} b^{3}) - \frac{d}{d a} (a^{3} b^{2}) - \frac{d}{d a} (b^{5})

\frac{d}{d a} (a^{5}) = 5 a^{4}

\frac{d}{d a} (a^{2} b^{3}) = 2 b^{3} a

\frac{d}{d a} (a^{3} b^{2}) = 3 b^{2} a^{2}

\frac{d}{d a} (b^{5}) = 0

= 5 a^{4} + 2 b^{3} a - 3 b^{2} a^{2} - 0

Simplify

= 5 a^{4} - 3 a^{2} b^{2} + 2 a b^{3}

\int \frac{2}{5} x^{\frac{5}{2}} d x

\frac{d x}{d t} = 4 x - x^{3}

\int_{0}^{\infty} x^{2} e^{- 2 x} d x

\int \frac{9 x ^{2}}{4 x - x ^{2}} d x

f (x) = 3 x + 91, (3, 10)

What is the d/(da)(a^5+a^2b^3-a^3b^2-b^5) ?
The d/(da)(a^5+a^2b^3-a^3b^2-b^5) is 5a^4-3a^2b^2+2ab^3
What is the first d/(da)(a^5+a^2b^3-a^3b^2-b^5) ?
The first d/(da)(a^5+a^2b^3-a^3b^2-b^5) is 5a^4-3a^2b^2+2ab^3