- #1
- 565
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How do I get the derivative of
y=tan4x
The answer in the back of the textbook
says (4/cos^2(4X))
y=tan4x
The answer in the back of the textbook
says (4/cos^2(4X))
i thought you would know [itex] \ sec^2x=\frac{1}{cos^2x} [/itex] :uhh:We haven't learned the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?