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How do you differentiate #ysinx=y#?

Answer 1

You can use the product rule, like the usual. Just remember that the derivative of y with respect to x in #(df)/(dx)# is #(dy)/(dx)#.

With #f(x): ysinx = y#,

#(df)/(dx)[ysinx = y] = ycosx + sinx((dy)/(dx)) = (dy)/(dx)#

#(dy)/(dx)[1-sinx] = ycosx#

#(dy)/(dx) = (ycosx)/(1-sinx)#

If you check Wolfram Alpha, you'll see this, just multiplied by #-1#.

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Answer 2

Aurora Alvarado

To differentiate the equation ysinx = y with respect to x, you would use the product rule. The product rule states that if you have two functions, u(x) and v(x), then the derivative of their product is given by (u'v + uv'). Applying this to the equation ysinx = y, where u(x) = y and v(x) = sinx, the derivative is:

(dy/dx)(sinx) + y(cosx) = dy/dx

Solving for dy/dx gives:

dy/dx = y(cosx) - y'(sinx)

This is the differentiation of ysinx = y with respect to x.

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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.