Differentiate?
#F(y) = (1/y^2 - 5/y^4) (y +7y^3) #

Question

Differentiate? 
#F(y) = (1/y^2 - 5/y^4) (y +7y^3) #

Evelyn Agard · Accepted Answer

#(dF(y))/dy = (-2y^-3+20y^-5)(y+7y^3)+(1/y^2-5/y^4)(1+21y^2)#

Apply the rule of the product:

#(dF(y))/dy = (d(1/y^2-5/y^4))/dy(y+7y^3)+(1/y^2-5/y^4)(d(y+7y^3))/dy#

Perform the initial derivative:

#(d(1/y^2-5/y^4))/dy=(d(y^-2-5y^-4))/dy =-2y^-3+20y^-5#

Return to the product rule by substituting:

#(dF(y))/dy = (-2y^-3+20y^-5)(y+7y^3)+(1/y^2-5/y^4)(d(y+7y^3))/dy#

Compute the second derivative:

#(d(y+7y^3))/dy = 1+21y^2#

Return to the product rule by substituting:

#(dF(y))/dy = (-2y^-3+20y^-5)(y+7y^3)+(1/y^2-5/y^4)(1+21y^2)#

Ezra Adams · Answer

undefined To differentiate the function $ F(y) = \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (y + 7y^3) $, you can use the product rule and the power rule.

First, apply the product rule:
$$ F'(y) = \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (1 + 21y^2) + \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (y + 7y^3)' $$

Now, differentiate each term using the power rule:
$$ F'(y) = \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (1 + 21y^2) + \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (1 + 21y^2) $$

Finally, simplify the expression:
$$ F'(y) = \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (1 + 21y^2) + \left(\frac{1}{y^2} - \frac{5}{y^4}\right) \cdot (1 + 21y^2) $$

$$ F'(y) = \frac{1}{y^2} + 21 - \frac{5}{y^4} - \frac{105}{y^2} + \frac{1}{y^2} + 21 - \frac{5}{y^4} - \frac{105}{y^2} $$

$$ F'(y) = \frac{2}{y^2} + 42 - \frac{10}{y^4} - \frac{210}{y^2} $$

So, the derivative of $ F(y) $ with respect to $ y $ is $ F'(y) = \frac{2}{y^2} + 42 - \frac{10}{y^4} - \frac{210}{y^2} $.

Differentiate? #F(y) = (1/y^2 - 5/y^4) (y +7y^3) #