Question

If $y = \frac{1}{2}x^{1/5} - \frac{3}{5}x^5$, then $\frac{dy}{dx} = $ (A) $\frac{2}{5x^{4/5}} + \frac{15}{x^6}$ (B) $\frac{2}{5x^{4/5}} + \frac{15}{x^4}$ (C) $\frac{2}{5x^{4/5}} - \frac{3}{5x^4}$ (D) $\frac{2x^{1/5}}{5} + \frac{15}{x^6}$ (E) $\frac{2x^{1/5}}{5} - \frac{3}{5x^4}$

          If $y = \frac{1}{2}x^{1/5} - \frac{3}{5}x^5$, then $\frac{dy}{dx} = $ 
(A) $\frac{2}{5x^{4/5}} + \frac{15}{x^6}$
(B) $\frac{2}{5x^{4/5}} + \frac{15}{x^4}$
(C) $\frac{2}{5x^{4/5}} - \frac{3}{5x^4}$
(D) $\frac{2x^{1/5}}{5} + \frac{15}{x^6}$
(E) $\frac{2x^{1/5}}{5} - \frac{3}{5x^4}$

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If y = (1)/(2)x^1/5 - (3)/(5)x^5, then (dy)/(dx) =
(A) (2)/(5x^4/5) + (15)/(x^6)
(B) (2)/(5x^4/5) + (15)/(x^4)
(C) (2)/(5x^4/5) - (3)/(5x^4)
(D) (2x^1/5)/(5) + (15)/(x^6)
(E) (2x^1/5)/(5) - (3)/(5x^4)

Added by Vincent B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Tim Thornhill

Step 1

x^(1/5) + 24/5) Show more…

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(A^2 + 5x^175) / (D^2.x^(1/5) + 24/5) = E^(3^5) x 4

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