# Edexcel P3

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16 years ago
#1
How would you differentiate (tanX)^x ???

0
16 years ago
#2
Hi! It's me again! Can anyone do this question? Pls pls Help!!!
0
16 years ago
#3
Originally posted by Unregistered
How would you differentiate (tanX)^x ???

I doubt such an exercise would come up in your P3 exam. Most people when doing this would use the chain rule straight off and get the wrong answer like this:

y = u^x, so dy/du = x*u^(x-1)
u = tan(x), so du/dx = sec^2(x)

And thus by the chain rule:
dy/dx = (dy/du)*(du/dx)
= (x*u^(x - 1))*sec^2(x)
= (x*(tan(x))^(x - 1))*sec^2(x)

This is, as mentioned, incorrect. The reason why is that when you differentiate y = u^x with respect to u, you are also differentiating implicitly with respect to x, as u is a function of x (u = tan(x)). Therefore, you can't just treat the x in the u^x as a constant and you're right back where you started.

To do this differentiation exercise correctly, we need to rewrite our expression cleverly as follows:

y = (tan(x))^x = e^(ln(tan(x))*x)

We can now differentiate y easily with respect to u as follows:

y = e^u, so dy/du = e^u

And then u with respect to x, using the product (and chain) rule:

u = ln(tan(x))x, so du/dx = ln(tan(x)) + x*cosec(x)sec(x)

Thus, using the chain rule again:

dy/dx = (dy/du)*(du/dx)
= e^u(ln(tan(x)) + x*cosec(x)sec(x))
= e^(ln(tan(x))x)*(ln(tan(x)) + x*cosec(x)sec(x))
= ((tan(x))^x)*(ln(tan(x)) + x*cosec(x)sec(x))

I hope that helps,

Kind Regards,
0
16 years ago
#4
Thank you so much! Really appreciate your help! 0
16 years ago
#5
Am currently trying to do P3 Edexcel from June 2001 and am stuck on the last question...

x=3 cos t
y=9 sin 2t

Find the Cartesian equation of the curve in the form y^2=f(x)

Anyone know how to do it?
0
16 years ago
#6
Originally posted by stuck!
Am currently trying to do P3 Edexcel from June 2001 and am stuck on the last question...

x=3 cos t
y=9 sin 2t

Find the Cartesian equation of the curve in the form y^2=f(x)

Anyone know how to do it?
x = 3cos(t)

y = 9sin(2t) = 18sin(t)cos(t) (standard trig. identity)

We know that cos(t) = x/3 from the first equation, so:

y = 18sin(t)cos(t) = 18sin(t)(x/3) = 6xsin(t)

We can now write them as:

cos(t) = x/3
sin(t) = y/(6x)

Using pythagoras:

sin^2(t) + cos^2(t) = 1
(y/(6x))^2 + (x/3)^2 = 1
(y^2)/(36*x^2) + x^2/9 = 1
y^2 = 36*x^2(1 - x^2/9)
= 36*x^2 - 4*x^4

Regards,
0
16 years ago
#7
Thank you!!
0
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