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How do you find #(dy)/(dx)# given #sqrt(3x^7+y^2)=x#?

Answer 1

Christopher Allers

#sqrt((3x^7 + y^2))^2 = x^2#

#3x^7 + y^2 = x^2#

#21x^6 + 2y(dy/dx) = 2x#

#2y(dy/dx) = 2x - 21x^6#

#dy/dx =(2x- 21x^6)/(2y)#

However, since #x = 3x^7 + y^2#, we can substitute.

#dy/dx = (2sqrt(3x^7 + y^2) - 21x^6)/(2y)#

Finally, we must note our restrictions on the variable. The #√# can never be negative, so #3x^7 + y^2 ≥ 0#. Also, #y !=0#.

Hopefully this helps!

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Answer 2

Charles Anctil

To find (dy)/(dx) given sqrt(3x^7+y^2) = x, you would implicitly differentiate both sides with respect to x and solve for (dy)/(dx).

The result is:

(dy)/(dx) = (-7x^6y)/(2sqrt(3x^7+y^2)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.