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How do you find #(dy)/(dx)# given #siny=x^2#?

Answer 1

Paisley Johnson

#y'(x) = (2x)/sqrt(1-x^4)#

Differentiate the equation with respect to #x#:

#siny = x^2#

#y' cosy = 2x#

#y' = (2x) / cosy#

If we restrict the function to the intervals #x in [0,1]# and #y in [0,pi/2]#:

#cosy = sqrt( 1-sin^2y) = sqrt ( 1-x^4)#

so that:

#y'(x) = (2x)/sqrt(1-x^4)#

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Answer 2

William Abdalla

To find (dy)/(dx) given siny = x^2, differentiate both sides of the equation implicitly with respect to x and solve for (dy)/(dx).

The derivative of siny with respect to x is cos(y) * (dy)/(dx), and the derivative of x^2 with respect to x is 2x.

So, we have:

cos(y) * (dy)/(dx) = 2x

Now, solve for (dy)/(dx):

(dy)/(dx) = 2x / cos(y)

Keep in mind that the expression (dy)/(dx) will be in terms of both x and y.

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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.