How do you differentiate #y=tanx/(2x^3)#?

Answer 1

#dy/dx = ( x^2sec^2x - 3tanx ) / (2x^4)#

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #, or less formally, # (u/v)' = (v(du)-u(dv))/v^2 #

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with # y=(tanx)/(2x^3) # Then
# { ("Let "u=tanx, => , (du)/dx=sec^2x), ("And "v=2x^3, =>, (dv)/dx=6x^2 ) :}#
# :. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 # # :. " " dy/dx = ( (2x^3)(sec^2x) - (tanx)(6x^2) ) / (2x^3)^2# # :. " " dy/dx = ( 2x^3sec^2x - 6x^2tanx ) / (4x^6)# # :. " " dy/dx = ( x^2sec^2x - 3tanx ) / (2x^4)#
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Answer 2

To differentiate ( y = \frac{\tan(x)}{2x^3} ), you can use the quotient rule. The quotient rule states that if ( y = \frac{u}{v} ), where ( u ) and ( v ) are functions of ( x ), then the derivative of ( y ) with respect to ( x ) is given by:

[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} ]

So, applying the quotient rule to the given function:

[ u = \tan(x) ] [ \frac{du}{dx} = \sec^2(x) ]

[ v = 2x^3 ] [ \frac{dv}{dx} = 6x^2 ]

Now, plug these into the quotient rule formula:

[ \frac{dy}{dx} = \frac{(2x^3)(\sec^2(x)) - (\tan(x))(6x^2)}{(2x^3)^2} ]

[ \frac{dy}{dx} = \frac{2x^3\sec^2(x) - 6x^2\tan(x)}{4x^6} ]

[ \frac{dy}{dx} = \frac{2\sec^2(x) - 6x\tan(x)}{4x^4} ]

This is the derivative of ( y ) with respect to ( x ).

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Answer 3

To differentiate ( y = \frac{\tan(x)}{2x^3} ) with respect to ( x ), you can use the quotient rule for differentiation:

If ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ),

where ( u' ) is the derivative of ( u ) with respect to ( x ), and ( v' ) is the derivative of ( v ) with respect to ( x ).

Applying the quotient rule to ( y = \frac{\tan(x)}{2x^3} ), we get:

( u = \tan(x) ) and ( v = 2x^3 ),

( u' = \sec^2(x) ) and ( v' = 6x^2 ).

Now, plug these values into the quotient rule formula:

( y' = \frac{\sec^2(x) \cdot 2x^3 - \tan(x) \cdot 6x^2}{(2x^3)^2} ).

Simplify this expression to get the derivative of ( y ) with respect to ( x ).

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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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