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Popular >

(-(q^4)/4+(2q^3)/3+(11q^2)/2-12q)^'

Solution

(- \frac{q ^{4}}{4} + \frac{2 q ^{3}}{3} + \frac{11 q ^{2}}{2} - 12 q)^{^{'}}

Solution

- q^{3} + 2 q^{2} + 11 q - 12

Solution steps

(- \frac{q ^{4}}{4} + \frac{2 q ^{3}}{3} + \frac{11 q ^{2}}{2} - 12 q)^{'}

Apply the Sum/Difference Rule:

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

= - (\frac{q ^{4}}{4})^{^{'}} + (\frac{2 q ^{3}}{3})^{^{'}} + (\frac{11 q ^{2}}{2})^{^{'}} - (12 q)^{^{'}}

(\frac{q ^{4}}{4})^{'} = q^{3}

(\frac{2 q ^{3}}{3})^{'} = 2 q^{2}

(\frac{11 q ^{2}}{2})^{'} = 11 q

(12 q)^{'} = 12

= - q^{3} + 2 q^{2} + 11 q - 12

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3 e^{6 x} \frac{d y}{d x} = - 36 \frac{x}{y ^{2}}

integral from 1/5 to 7 of 6xln(5x)

\int_{\frac{1}{5}}^{7} 6 x ln (5 x) d x

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f (x) = (x^{2} - 9)^{2}

(dy)/(dx)=(3y^2cot(x)+sin(x)cos(x))/(2y)

\frac{d y}{d x} = \frac{3 y ^{2} cot ( x ) + sin ( x ) cos ( x )}{2 y}

limit as x approaches 10 of 8/((x-10)^2)

x \to 10 lim (\frac{8}{( x - 10 ) ^{2}})

Frequently Asked Questions (FAQ)

What is the (-(q^4)/4+(2q^3)/3+(11q^2)/2-12q)^' ?
The (-(q^4)/4+(2q^3)/3+(11q^2)/2-12q)^' is -q^3+2q^2+11q-12
What is the first (-(q^4)/4+(2q^3)/3+(11q^2)/2-12q)^' ?
The first (-(q^4)/4+(2q^3)/3+(11q^2)/2-12q)^' is -q^3+2q^2+11q-12