How do you find the derivative of # (3+sin(x))/(3x+cos(x))#?

Answer 1

Use quotient rule to obtain #(f/g)'=(3xcosx-8)/(9x^2+6xcosx+cos^2x)#

Quotient rule: #(f/g)'=(f'g-g'f)/g^2# Let #f=3+sinx#; therefore #f'=cosx# Let #g=3x+cosx#; therefore #g'=3-sinx# #(f/g)'=(f'g-g'f)/g^2# #=[cosx(3x+cosx)-(3-sinx)(3+sinx)]/(3x+cosx)^2# #=[(3xcosx+cos^2x)-(9-sin^2x)]/(9x^2+6xcosx+cos^2x)# #=(3xcosx+sin^2x+cos^2x-9)/(9x^2+6xcosx+cos^2x)# Identity: #sin^2x+cos^2x=1=>(f/g)'=(3xcosx-8)/(9x^2+6xcosx+cos^2x)#

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Answer 2

Adam Adkins

To find the derivative of ( \frac{3+\sin(x)}{3x+\cos(x)} ), you can use the quotient rule, which states that for functions ( u(x) ) and ( v(x) ), the derivative of ( \frac{u(x)}{v(x)} ) is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ). Applying this rule to the given function, the derivative is:

[ \frac{(3+\sin(x))(3x+\cos(x))' - (3x+\cos(x))(3+\sin(x))'}{(3x+\cos(x))^2} ]

[ = \frac{(3+\sin(x))(3-\sin(x)) - (3x+\cos(x))(3+\cos(x))}{(3x+\cos(x))^2} ]

[ = \frac{9 - \sin^2(x) - 9x\cos(x) - 3\sin(x) - 3\cos(x) - x\cos^2(x)}{(3x+\cos(x))^2} ]

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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.