Help with C3 Differentiation question

Hi,

I have no idea how to work out this question, and the mark scheme has skipped the step I need.

The question is as follows:

7b) Given that x = 3tan2y find dy/dx in terms of x.

My progress:
Dx/dy = 6Sec^2(2y)
Dy/dx = 1/6sec^2(2y)

I have no idea where to go from here, so I appreciate any help.
The answer should be 3/(18+2x^2).

Reply 1

B_9710

18

Original post by Jamdroid

Hi,

I have no idea how to work out this question, and the mark scheme has skipped the step I need.

The question is as follows:

7b) Given that x = 3tan2y find dy/dx in terms of x.

My progress:
Dx/dy = 6Sec^2(2y)
Dy/dx = 1/6sec^2(2y)

I have no idea where to go from here, so I appreciate any help.
The answer should be 3/(18+2x^2).

sec^2(2y)=tan^2(2y)+1
How could you write this in terms of x?

Reply 2

B_9710

18

Alternatively you could say that y=(1/2)arctan(x/3) and then differentiate both sides with respect to x, this is assuming you know how to differentiate the inverse trig functions. If not then stick to the first method.

Reply 3

Jamdroid

OP

2

Original post by B_9710

sec^2(2y)=tan^2(2y)+1
How could you write this in terms of x?

So...
Tan^2(2y) = (x^2)/9
Sec^2(2y) = tan^2(2y) + 1

Dx/dy = 6Sec^2(2y)
= 6((x^2)/9 + 1)
= (6x^2)/9 + 6
= (2x^2)/3 + 18/3

Dy/dx = 3/(18+2x^2)

Makes perfect sense, I don't know how I didn't think of it!

Maybe I'll learn how to differentiate inverse functions too, why not?

Thank you for your help sir, I really appreciate it!

Help with C3 Differentiation question

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