How do you differentiate #f(x)=4x*(x-4)*tanx# using the product rule?
Simplifying by factoring:
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# f'(x) = 8(x-2)tanx + 4x(x-4)sec^2x #
We have:
We will apply the Product Rule for Differentiation:
I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".
This can be extended to three products:
Then:
Gives us:
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To differentiate ( f(x) = 4x(x - 4)\tan(x) ) using the product rule:
- Identify the two functions being multiplied: ( u(x) = 4x(x - 4) ) and ( v(x) = \tan(x) ).
- Apply the product rule formula: ( (uv)' = u'v + uv' ).
- Differentiate each function: ( u'(x) ) (derivative of ( u(x) )) = ( (4x)'(x - 4) + 4x(x - 4)' ) ( v'(x) ) (derivative of ( v(x) )) = ( \sec^2(x) ) (derivative of ( \tan(x) )).
- Substitute the derivatives and the original functions into the product rule formula. ( f'(x) = (4(x - 4) + 4x) \tan(x) + 4x(x - 4) \sec^2(x) ).
- Simplify the expression, if necessary, to get the final result.
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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.
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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