Calculate y'

[Dazed]

Could someone please help me solve these step by step? Thanks.

Calculate y'

1. y = 1/sin(x-sinx)

2. y = sec(1+x^2)

3. x^2cosy + sin2y = xy

4. y = sinmx/x

5. xtany = y-1

6. y = (x-1)(x-4)/(x-2)(x-3)

 

[ChaoticLlama]

y = sinmx/x

First I will bring the denominator into the numerator to make it a product rule.
y = sin(mx) * (x^-1)

Then Differentiate using product rule.
y' = sin(mx) * -(x^-2) + (x^-1)cos(mx)m
y' = -sin(mx)/x² + mcos(mx)/x
y' = (mxcos(mx) - sin(mx))/x²

y = sec(1 + x^2)

Since the derivative of sec(x) = sec(x)tan(x)

y' = sec(1 + x²)tan(1 + x²) * 2x

 

[ChaoticLlama]

xtany = y-1

Here you must make use of product rule on the right side and implicit differentiation.

x * sec²(y) * y' + tan(y) = y'

Then just rearragnge to solve for y'

x * sec²(y) * y' - y' = -tan(y)

y'(xsec²(y) - 1) = -tan(y)

y' = -tan(y) / [xsec(y) - 1]


y = 1/sin(x-sinx)

Here, all that you have to keep in mind is that the derivative of sec(u) = sec(u)tan(u) * du/dx
The du/dx is not officially a part of the derivative of sec(x), it is just there to remind you of chain rule. (where u = x - sin(x))


y = sec(x - sin(x))

y' = sec(x-sin(x))tan(x - sin(x))(1 - cos(x))