# How do you differentiate #(sinx+sinxcosx)/x#?

we need the quotient rule and the product rule

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To differentiate the expression (sinx + sinx*cosx)/x, you can use the quotient rule of differentiation. The quotient rule states that if you have a function in the form of u(x)/v(x), then the derivative of this function with respect to x is (v(x)*u'(x) - u(x)*v'(x)) / [v(x)]^2. Applying this rule to the given expression:

Let u(x) = sinx + sinx*cosx and v(x) = x.

Then, u'(x) = cosx + (cosx*cosx - sinx*sinx) = cosx + cos^2(x) - sin^2(x)

And v'(x) = 1

Now, applying the quotient rule:

[(x)(cosx + cos^2(x) - sin^2(x)) - (sinx + sinx*cosx)(1)] / [x]^2

Expanding and simplifying:

(cosx*x + cos^2(x)*x - sin^2(x)*x - sinx - sinx*cosx) / x^2

(cosx*x + x*cos^2(x) - x*sin^2(x) - sinx - sinx*cosx) / x^2

(cosx*x + x*cos^2(x) - x*sin^2(x) - sinx - sinx*cosx) / x^2

(cosx + x*cos^2(x) - x*sin^2(x) - sinx - sinx*cosx) / x

(cosx + x*cos^2(x) - x*sin^2(x) - sinx - sinx*cosx) / x

(cosx - sinx - sinx*cosx) / x

So, the derivative of (sinx + sinx*cosx)/x with respect to x is (cosx - sinx - sinx*cosx) / x.

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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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