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How do you find the derivative of #f(x)=( 6tanx-10)/secx#?

Answer 1

Isabella Ackerman

#f'(x)=6cosx+10sinx#

We can make this simpler by first simplifying the function:

#f(x)=(6tanx-10)/secx=(6(sinx/cosx)-10)/(1/cosx)#

#f(x)=cosx(6sinx/cosx-10)=6sinx-10cosx#

Now, all we need to know is that #d/dxsinx=cosx# and #d/dxcosx=-sinx#. Then:

#f'(x)=6cosx-10(-sinx)=6cosx+10sinx#

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Answer 2

Charles Anctil

To find the derivative of ( f(x) = \frac{6\tan x - 10}{\sec x} ), you can use the quotient rule.

The quotient rule states that if you have a function ( u(x) ) divided by ( v(x) ), the derivative is given by:

[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

So, applying the quotient rule to ( f(x) ):

[ f'(x) = \frac{(6(\sec^2 x) - 6\tan x \cdot \sec x) - ((6\tan x - 10)(\sec x \tan x))}{(\sec x)^2} ]

Simplify this expression to get the derivative of ( f(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.