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NEED help differentation

will post my question in a min
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Qer
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i am watching a tutorial on exam solutions . i stuck here in this tutorial he is using a chain rule but why he wrote 7 and 3 . i dont understand this can anyone please explain this to me .
Reply 2
Avatar for MR1999
MR1999
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Original post by Qer
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i am watching a tutorial on exam solutions . i stuck here in this tutorial he is using a chain rule but why he wrote 7 and 3 . i dont understand this can anyone please explain this to me .


You need to apply the chain rule twice. So for y=4(tan⁡3x)5, \displaystyle y=4(\tan 3x)^5, let u=3xu=3x and v=tan⁡u. v=\tan u.

Then dydx=dydv⋅dvdu⋅dudx\displaystyle \frac{dy}{dx} = \frac{dy}{dv}\cdot \frac{dv}{du}\cdot \frac{du}{dx}
(edited 6 years ago)
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yusyus
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Original post by Qer
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i am watching a tutorial on exam solutions . i stuck here in this tutorial he is using a chain rule but why he wrote 7 and 3 . i dont understand this can anyone please explain this to me .


try and differentiate a trig value with a constant using a substitution it might help
Original post by Qer
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i am watching a tutorial on exam solutions . i stuck here in this tutorial he is using a chain rule but why he wrote 7 and 3 . i dont understand this can anyone please explain this to me .

This is where you have to use the chain rule twice.

ddx[4(tan(3x))5]=5×4tan(3x)4×ddx(tan(3x)) \frac{d}{dx} [4(tan(3x))^5] = 5 \times4 tan(3x)^4 \times \frac{d}{dx}(tan(3x))
Which I think you understand.

Now you have to use the chain rule again as ddx(tan(3x))=3sec2(3x)\frac{d}{dx}(tan(3x)) = 3sec^2(3x)
(edited 6 years ago)
Reply 5
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Qer
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again get confused

so 4tan^5 (3x)
4[tan 3x]^5

20[sec^2 3x] tan 3x]^4

tell whereto used chain rule now
(edited 6 years ago)
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Qer
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@NotNotBatman
@bruh2132
@Desmos
Reply 7
Avatar for MR1999
MR1999
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Original post by Qer
again get confused

so 4tan^5 (3x)
4[tan 3x]^5

20[sec^2 3x] tan 3x]^4

tell whereto used chain rule now


Apply chain rule to tan3x
Reply 8
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Qer
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Original post by Desmos
Apply chain rule to tan3x


so 4tan^5 (3x)
4[tan 3x]^5

20[sec^2 3x] tan 3x]^4

applying chain rule to [tan3x ]^4= 4(sec^2 3x] [tan3x]^3
like that????
Reply 9
Avatar for MR1999
MR1999
21
Original post by Qer
so 4tan^5 (3x)
4[tan 3x]^5

20[sec^2 3x] tan 3x]^4

applying chain rule to [tan3x ]^4= 4(sec^2 3x] [tan3x]^3
like that????


No just to tan⁡3x \tan 3x. Let u=3x u = 3x and y=tan⁡uy=\tan u. Then y′(x)=u′(x)sec⁡2uy'(x) =u'(x) \sec^2 u

When you differentiate a composite function, you must differentiate every function which makes it up, whether it's constant, linear, quadratic, trig, exponential etc.
(edited 6 years ago)

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