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How do you find the derivative of #f(x)= (5x^2 tan(x))/sec(x)#?

Answer 1

#f'(x)=5x^2cos(x)+10xsin(x)#

Rewrite #tan(x)# as #sin(x)/cos(x)# and #1/sec(x)# as #cosx#. As you can see, both the cos(x) cancel out leaving you with a simpler function to solve for.

#f(x)=5x^2sin(x)#

From here, use the product rule. Note that the derivative of #sinx# is #cosx#:

#f'(x)=5x^2cos(x)+sin(x)(10x)#

Simplify:

#f'(x)=5x^2cos(x)+10xsin(x)#

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Answer 2

Charles Arocha

To find the derivative of ( f(x) = \frac{5x^2 \tan(x)}{\sec(x)} ), use the quotient rule:

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

where ( u = 5x^2 \tan(x) ) and ( v = \sec(x) ). Differentiate both ( u ) and ( v ) with respect to ( x ):

[ u' = 10x \tan(x) + 5x^2 \sec^2(x) ] [ v' = \sec(x) \tan(x) ]

Then plug these into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{5x^2 \tan(x)}{\sec(x)} \right) = \frac{\sec(x)(10x \tan(x) + 5x^2 \sec^2(x)) - 5x^2 \tan(x) \sec(x) \tan(x)}{\sec^2(x)} ]

Simplify this expression to get the derivative.

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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.