C3 differentiation !!!!! :(

ok, so I thought I was understanding differentiation fine until I came across this question..

2tan^4 (5x- pi) I changed it to 2tan(5x-pi)^4 so my dy/dx= 8sec^2 (5x-pi)^3 x (5)= 40sec^2 (5x-pi), where am I going wrong ??

Please help a v.confused student

ok, so I thought I was understanding differentiation fine until I came across this question..

2tan^4 (5x- pi) I changed it to 2tan(5x-pi)^4 so my dy/dx= 8sec^2 (5x-pi)^3 x (5)= 40sec^2 (5x-pi), where am I going wrong ??

Please help a v.confused student

2tan⁴ (5x- pi) change it to 2[ tan(5x-pi) ]<sup>4

try it like this</sup>

How many marks is it worth?
Have you covered double angle formulae?

EDIT: scratch that - it would just complicate things

y = 2[tan( 5x - π )]⁴




let u = tan( 5x - π )

du/dx = 5sec²( 5x - π )

y =2u⁴


dy/du = 8u³ = 8tan³( 5x - π )

dy/dx = dy/dy*du/dx....

ok, so I thought I was understanding differentiation fine until I came across this question..

2tan^4 (5x- pi) I changed it to 2tan(5x-pi)^4 so my dy/dx= 8sec^2 (5x-pi)^3 x (5)= 40sec^2 (5x-pi), where am I going wrong ??

Please help a v.confused student

Ohh ok, the extra bracket makes it look a lot more neater. I'll give it a go.




no idea- its a question from an exam solutions vid.



do I need to continue with the differentiation or is this the final answer ?

Be carful with using chain rule correctly.

You should have

$\dfrac{\mathrm{d} y}{\mathrm{d} x } = 8 \tan^3 (5x - \pi)^3 \cdot \dfrac{\mathrm{d}}{\mathrm{d} x } \bigg[ \tan{(5x - \pi)} \bigg]$

I'm so confused omg

How would I know to differentiate this twice???

Lmao so am I and I did a maths degree (well half)

I need a pen and paper to work it out Lol.

Well you use the chain rule, for a composite function the differential is given by

$f'(g(x)) \cdot g'(x)$


So in your example $f(x) = 2x^4$ and $g(x) = \tan{(5x - \pi)}$

To get a final answer you would need to find g'(x) (which I left up to you).

I'm so confused omg

I know it feels confusing however just think about it logically. It is tricky because you need to use the chain rule twice (since $\dfrac{\mathrm{d} }{\mathrm{d} x} \tan{(5x - \pi)}$ requires chain rule).

My differentiation confidence is slowly declining..




I seriously do not know how to add the sec^2 onto it. I just need a step by step clear method

I'm so confused omg

y = 2[tan( 5x - π )]⁴

let u = tan( 5x - π )

du/dx = 5sec²( 5x - π )

y =2u⁴

dy/du = 8u³ = 8tan³( 5x - π )

dy/dx = dy/dy*du/dx....

Sec comes into it because you need to multiply by $g'(x)$ .

Evaluate $\dfrac{\mathrm{d} }{\mathrm{d} x} \bigg[ \tan{(5x - \pi)} \bigg]$

and you should see why sec is part of the answer.

Hey, look this guy did it here above, he used the chain rule.

Check this video out if you're still confused, there is a very similar example that Mr examsolutions goes through: http://www.examsolutions.net/maths-revision/core-maths/differentiation/methods/chain-rule/sincostan%5En/tutorial-1.php

gl

It's the same question lol, I did watch the vid. I'm still stuck