SOLUTION: Obtain dy/dx (rate of change) using the direct rule:
y= 2.5x^4 - x^3/2.5 - x/2.5 +2.5
&
y= (x^3-2.5x^2)/(2.5sqrtx)
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Question 343519: Obtain dy/dx (rate of change) using the direct rule:
y= 2.5x^4 - x^3/2.5 - x/2.5 +2.5
&
y= (x^3-2.5x^2)/(2.5sqrtx)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I'm unfamiliar with the term "direct rule." For the first function you need repeated application of the power rule (presuming that I am interpreting your rendering correctly)
Power Rule
So:
\ =\ 10x^3\ -\ \frac{6x^2}{5}\ -\ \frac{2}{5})
I'm going to do the second one for you as well because I'm in a good mood. But next time -- READ THE INSTRUCTIONS -- One question per post.
For the second one you need the Quotient Rule and the Power Rule:
Quotient Rule
For:
\ =\ \frac{f(x)}{g(x)})
where
\ \ )
and \ \neq\ 0)
are both differentiable,
\ =\ \frac{f'(x)g(x)\ -\ f(x)g'(x)}{\left[g(x)\right]^2})
And we want to calculate:
Begin with calculating the derivatives of numerator and denominator separately:
\ =\ x^3\ -\ 2.5x^2)
hence
\ =\ 3x^2\ -\ 5x)
and
\ =\ 2.5\sqrt{x}\ =\ 2.5x^{\frac{1}{2})
So
\ =\ 1.25x^{-\frac{1}{2}}\ =\ \frac{1.25}{x^{\frac{1}{2}})
\left(2.5x^{\frac{1}{2}}\right)\right]\ -\ \left[\left(x^3\ -\ 2.5x^2\right)\left(\frac{1.25}{x^{\frac{1}{2}}}\right)\right]}{\left[2.5x^{\frac{1}{2}}\right]^2}\ =\ \frac{\left(7.5x^{\frac{5}{2}}\ -\ 12.5x^{\frac{3}{2}}\right)\ -\ \left(1.25x^{\frac{5}{2}}\ -\ 3.125x^{\frac{3}{2}}\right)}{6.25x})
\ -\ \left(x^{\frac{3}{2}}\ -\ 2.5x^{\frac{1}{2}}\right)}{5})
John
My calculator said it, I believe it, that settles it
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