How do you differentiate # g(x) = (cotx + cscx)/(tanx - sinx) #?

Answer 1

#(dg)/(dx)=-cotxcot^2(x/2)(csc(x/2)+cscx)#

Quotient rule states if #g(x)=(a(x))/(b(x))#
then #(dg)/(dx)=((da)/(dx)xxb(x)-(db)/(dx)xxa(x))/(b(x))^2#
Here #a(x)=cotx+cscx# and #b(x)=tanx-sinx#
i.e. #(da)/(dx)=-csc^2x-cotxcscx# and #(db)/(dx)=sec^2x-cosx#
Hence #(dg(x))/(dx)=([(tanx-sinx)(-csc^2x-cotxcscx)-(sec^2x-cosx)(cotx+cscx)])/(tanx-sinx)^2#

= #([-tanxcsc^2x-cscx+cscx+cotx -secxcscx-sec^2xcscx+cosxcotx+cotx])/(tanx-sinx)^2#

= #([-tanxcsc^2x+2cotx -secxcscx-sec^2xcscx+cosxcotx])/(tanx-sinx)^2#

We can simplify it further by converting each trigonometric ratio into #sinx# and #cosx# and simplifying it further
but the simpler way could be that as #g(x)=(cotx+cscx)/(tanx-sinx)#
= #(cosx/sinx+1/sinx)/(sinx/cosx-sinx)#
= #((cosx+1)/sinx)/((sinx-sinxcosx)/(sinxcosx)#
= #(cosx(1+cosx))/(sinx(1-cosx))#
=#cotxcot^2(x/2)#
and hence #(dg)/(dx)=2cotxcot(x/2)(-cot(x/2)csc(x/2))xx1/2+cot^2(x/2)(-cotxcscx)#
= #-cotxcot^2(x/2)csc(x/2)-cotxcscxcot^2(x/2)#
= #-cotxcot^2(x/2)(csc(x/2)+cscx)#
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Answer 2

To differentiate ( g(x) = \frac{\cot(x) + \csc(x)}{\tan(x) - \sin(x)} ), follow these steps:

  1. Express all trigonometric functions in terms of sine and cosine.
  2. Apply the quotient rule for differentiation.
  3. Simplify the expression.

Here are the detailed steps:

  1. Rewrite cotangent, cosecant, and tangent functions in terms of sine and cosine:

    • ( \cot(x) = \frac{\cos(x)}{\sin(x)} )
    • ( \csc(x) = \frac{1}{\sin(x)} )
    • ( \tan(x) = \frac{\sin(x)}{\cos(x)} )
  2. Apply the quotient rule:

    • Let ( u(x) = \cot(x) + \csc(x) ) and ( v(x) = \tan(x) - \sin(x) ).
    • Then, ( g(x) = \frac{u(x)}{v(x)} ).
    • Differentiate ( u(x) ) and ( v(x) ) separately.
    • Use the quotient rule: ( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} ).
  3. Simplify the expression:

    • Substitute the derivatives of ( u(x) ) and ( v(x) ) into the quotient rule expression.
    • Simplify the resulting expression.

The final expression after differentiation and simplification would be the derivative of ( g(x) ).

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Answer 3

To differentiate the function g(x) = (cot(x) + csc(x))/(tan(x) - sin(x)), you can use the quotient rule:

Let u = cot(x) + csc(x) and v = tan(x) - sin(x).

Then, g'(x) = (u'v - uv') / v^2, where u' and v' are the derivatives of u and v respectively.

Now, find the derivatives: u' = -csc^2(x) - csc(x)cot(x) v' = sec^2(x) - cos(x)

Plug these values into the quotient rule formula:

g'(x) = ((-csc^2(x) - csc(x)cot(x))(tan(x) - sin(x)) - (cot(x) + csc(x))(sec^2(x) - cos(x))) / (tan(x) - sin(x))^2

Simplify the expression if necessary.

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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.

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.

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.

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.

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