# How do you differentiate #y=csctheta(theta+cottheta)#?

we will need to use the product rule

tiding up.

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To differentiate ( y = csc(\theta) (\theta + cot(\theta)) ), we can use the product rule and the derivatives of the cosecant and cotangent functions:

Product Rule: ( (uv)' = u'v + uv' )

Derivative of cosecant: ( \frac{d}{d\theta} csc(\theta) = -csc(\theta) cot(\theta) )

Derivative of cotangent: ( \frac{d}{d\theta} cot(\theta) = -csc^2(\theta) )

Now, let's differentiate ( y ):

[ y' = \frac{d}{d\theta} \left[csc(\theta) (\theta + cot(\theta))\right] ]

[ = \frac{d}{d\theta} [csc(\theta)] (\theta + cot(\theta)) + csc(\theta) \frac{d}{d\theta} (\theta + cot(\theta)) ]

[ = (-csc(\theta) cot(\theta))(\theta + cot(\theta)) + csc(\theta)(1 - csc^2(\theta)) ]

[ = -csc(\theta) \cdot cot(\theta) \cdot \theta - csc(\theta) \cdot cot^2(\theta) + csc(\theta) - csc^3(\theta) ]

[ = -\theta csc(\theta) cot(\theta) - csc(\theta) \cdot \frac{cos^2(\theta)}{sin^2(\theta)} + csc(\theta) - \frac{1}{sin^3(\theta)} ]

[ = -\theta csc(\theta) cot(\theta) - \frac{cos^2(\theta)}{sin(\theta)} + csc(\theta) - \frac{1}{sin^3(\theta)} ]

[ = -\theta csc(\theta) cot(\theta) - \frac{1 - sin^2(\theta)}{sin(\theta)} + csc(\theta) - \frac{1}{sin^3(\theta)} ]

[ = -\theta csc(\theta) cot(\theta) - \frac{1}{sin(\theta)} + \frac{sin(\theta)}{sin(\theta)} + csc(\theta) - \frac{1}{sin^3(\theta)} ]

[ = -\theta csc(\theta) cot(\theta) - \frac{1}{sin(\theta)} + \frac{1}{sin(\theta)} + csc(\theta) - \frac{1}{sin^3(\theta)} ]

[ = -\theta csc(\theta) cot(\theta) + csc(\theta) - \frac{1}{sin^3(\theta)} ]

So, the derivative of ( y ) with respect to ( \theta ) is:

[ y' = -\theta csc(\theta) cot(\theta) + csc(\theta) - \frac{1}{sin^3(\theta)} ]

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