Show that the derivative of tan 3x = 3Sec^2 3x

U

ugu

Junior Member
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Nov 23, 2021
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QUESTION

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Solution : i will appreciate correction when necessary

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I don't understand the first symbol in the 3rd line, and the way I read the 4th line is [imath]\Rightarrow \frac{dy}{d\mathbf{x}} = \sec^2u \;\;\; \frac{d\mathbf y}{dx}=3[/imath], which would have two typos. But otherwise it looks good to me.
 
blamocur said:
I don't understand the first symbol in the 3rd line, and the way I read the 4th line is [imath]\Rightarrow \frac{dy}{d\mathbf{x}} = \sec^2u \;\;\; \frac{d\mathbf y}{dx}=3[/imath], which would have two typos. But otherwise it looks good to me.
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it is let y =. and yes, what you stated is what was written
 
ugudansam said:
i need them fixed; as i am no too good with at all
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If you are not too good with the subject we can help you to get better, instead of simply correcting your mistakes. For starters: why do you think that [imath]\frac{dy}{dx} = \sec^2 u[/imath]?
 
the truth is i do not know, i was just trying to follow some example to solve this. I am not yet getting along with this cause of differential and integral calculus as well as Trigonometry and Coordinate geometry.
 
You have the chain rule spelled out in the second line, but it seems to have a typo too. The correct one is [imath]\frac{dy}{dx}=\frac{dy}{du} \frac{d\mathbf u}{dx}[/imath]. But your 3rd line is somewhat sloppy -- can you tell me what you mean there?
 
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