+26367 How do I solve this in hand: diff(f(x), x) = 4*x^3-4/x^2
I assume the question is: \(\dfrac{d\ \left( 4\cdot x^3 - \frac{4}{x^2 } \right) } {dx} = \ ?\)
\(\begin{array}{rcl} y &=& 4\cdot x^3 - \frac{4}{x^2 } \\ y &=& 4 \cdot \left( x^3 - \frac{1}{x^2} \right)\\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( \frac{1}{x^2} \right)' ] \\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( x^{-2} \right)' ] \\ \boxed{~ \begin{array}{lcl} \text{rule:} \qquad \left(x^n\right)' &=& n x^{n-1} \end{array} ~}\\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) - \left( (-2)\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) + \left( 2\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ 3x^{2} + 2\cdot x^{-3} ] \\ y' &=& 4 \cdot [ 3x^{2} + \frac{2}{x^3} ] \\ y' &=& 12x^{2} + \frac{8}{x^3} \\ \end{array}\)
It has been a while since I have seen Diffy Q, but if I recall correctly,
4x^3 - 4/x^2
4x^3 - 4x^-2
DIff:
12x2 + 8x^-3 ? Maybe YOU can help ME on this one !
+26367 How do I solve this in hand: diff(f(x), x) = 4*x^3-4/x^2
I assume the question is: \(\dfrac{d\ \left( 4\cdot x^3 - \frac{4}{x^2 } \right) } {dx} = \ ?\)
\(\begin{array}{rcl} y &=& 4\cdot x^3 - \frac{4}{x^2 } \\ y &=& 4 \cdot \left( x^3 - \frac{1}{x^2} \right)\\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( \frac{1}{x^2} \right)' ] \\ y' &=& 4 \cdot [ \left( x^3 \right)' - \left( x^{-2} \right)' ] \\ \boxed{~ \begin{array}{lcl} \text{rule:} \qquad \left(x^n\right)' &=& n x^{n-1} \end{array} ~}\\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) - \left( (-2)\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ \left( 3\cdot x^{3-1} \right) + \left( 2\cdot x^{-2-1} \right) ] \\ y' &=& 4 \cdot [ 3x^{2} + 2\cdot x^{-3} ] \\ y' &=& 4 \cdot [ 3x^{2} + \frac{2}{x^3} ] \\ y' &=& 12x^{2} + \frac{8}{x^3} \\ \end{array}\)
