DCTP Forums

Whatever's in the parantheses in the f(x) form needs to be substituted in for the x value of the equivalent fuction before it can be used.

As for that second question... You're supposed to set the x function equal to t, and then substitute that in to the t in the y function, and then differentiate normally.

Conan324 wrote: And if I'm asked to find derivative with respect to x, does the answer has to have an x in it?

It's actually telling you to write the derivative in d/dx form, and it doesn't have to have an x variable in it (but it should have an x differential (dx) in it). For example, the derivative of y=5t with respect to x would be dy/dx=5dt/dx.

the first one is easy

* = multiplicate


f(5x) = x³

f(5*x) = x * x * x    /  :x

                      __
f 5 = x²  /    √


f 2,23 = x

x= 2,23

2,23 * 5 = 11,18
2,23^3 = 11,18


ive come back from night work and maked this but i dont think i can remember the cos sin tan thing before sleeping

The method for the second one is correct. As is the answer. For the first one F(5x)=x^3
=> 5F'(5x) = 3x^2. put x=1 => F'(5) = 3/5.

Conan324 wrote: Thanks everyone for your valued inputs , as for the second question, i couldn't solve for T because we haven't been taught the derivatives of the inverse trigonometric functions so i had to use implicit differentiation.

but im not sure about the first question since i got mixed answers :X,i initially thought the method to be like what aly_angelflight suggested, but conan-chandesune provided any alternate method and i don't know which one is true ,  any further elaboration  would be much appreciated.

Actually aly_angelflight's method and my method are inherently similar. Both of us just went the other way around it.

I see some inconsistencies here so I'm just gonna explain things...

For the first question, Conan-chandesune's solution is right. Another way to see it is:
If f(5x)=x³
f(x) = f(5(x/5)) = x³/125
f'(x) = 3x²/125 --> f'(5) =  3/5

For the second question, I also thought the same as Aly's solution but since you mentioned that you didn't know the derivatives of inverse trigonometric functions yet, I thought of one that only requires implicit differentiation.

It may not be noticed easily but if,
y=cos(5t) and x=sin(5t)
x² + y² = sin²(5t) + cos²(5t) = 1

So your equation becomes...
x² + y² = 1
and you proceed to normal implicit differentiation. (won't show the solution since it's too long, and I believe you know how to do it)

Conan324 wrote:

Conan-chandesune wrote:

Conan324 wrote: Thanks everyone for your valued inputs , as for the second question, i couldn't solve for T because we haven't been taught the derivatives of the inverse trigonometric functions so i had to use implicit differentiation.

but im not sure about the first question since i got mixed answers :X,i initially thought the method to be like what aly_angelflight suggested, but conan-chandesune provided any alternate method and i don't know which one is true ,  any further elaboration  would be much appreciated.

Actually aly_angelflight's method and my method are inherently similar. Both of us just went the other way around it.

But I got different answers when using the two methods :s, I don't think I understand it yet...

Nope. You must have hyped up the calculations somewhere. As Miyano shiho illustrates, the answer is 3/5 using aly's method too.

Hyped??? XD

miyano_shiho wrote: Hyped??? XD

it's part of the DCTP word filter, he meant to say "f-ed up" :V