How do you differentiate $y = (\frac{8}{x^{3}}) (\frac{x + x^{7}}{5})$ $y= (8/x^3)((x + x^7)/5)$ ?

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Answer 1 · 2016-06-22T13:22:00

$\frac{d y}{d x} = - \frac{16}{5 x^{3}} + \frac{32}{5} x^{3}$ $(dy)/(dx)=-16/(5x^3)+32/5x^3$

The function $y = (\frac{8}{x^{3}}) (\frac{x + x^{7}}{5})$ $y=(8/x^3)((x+x^7)/5)$

= $\frac{8}{5} \times (\frac{x + x^{7}}{x^{3}})$ $8/5xx((x+x^7)/x^3)$

= $\frac{8}{5} \times (\frac{x}{x^{3}} + \frac{x^{7}}{x^{3}})$ $8/5xx(x/x^3+x^7/x^3)$

= $\frac{8}{5} \times (\frac{1}{x^{2}} + x^{4})$ $8/5xx(1/x^2+x^4)$

Hence $\frac{d y}{d x} = \frac{8}{5} \times (\frac{- 2}{x^{3}} + 4 x^{3})$ $(dy)/(dx)=8/5xx((-2)/x^3+4x^3)$

or $\frac{d y}{d x} = - \frac{16}{5 x^{3}} + \frac{32}{5} x^{3}$ $(dy)/(dx)=-16/(5x^3)+32/5x^3$

How do you differentiate y= (8/x^3)((x + x^7)/5)y=(8x3)(x+x75)y= (8/x^3)((x + x^7)/5)?