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# Differentiation help!

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1. May sound silly question, but trying to teach myself... can someone help me with finding the derivatives of the following equations:

e^x/2

Ln(x/4)

Find the gradient of the curve whose equation is y=e^3x at the point (0,1)
2. You should look into chain derivatives for compositions of functions. Otherwise, you have to rely upon a table of derivatives for the derivative of the natural logarithm.
If you ever need examples of full solutions to derivatives, you should try out this derivative calculator with steps.
3. (Original post by Olmeister)
May sound silly question, but trying to teach myself... can someone help me with finding the derivatives of the following equations:

e^x/2

Ln(x/4)

Find the gradient of the curve whose equation is y=e^3x at the point (0,1)
Use the chain rule.
4. As people above me said, use the chain rule... for the ln(x/4) though I would use one of the log rules first
5. (Original post by asinghj)
As people above me said, use the chain rule... for the ln(x/4) though I would use one of the log rules first
We want dy/dx
y=ln(x/4)
let u = x/4
du/dx = 1/4
...
y=ln(u)
dy/du = 1/u
1/u = 4/x

therefore dy/du *du/dx = dy/dx (in this case behaves like a fraction and du cancels)
so 4/x * 1/4 = 1/x
dy/dx = 1/x
6. (Original post by Kawaii289)
We want dy/dx
y=ln(x/4)
let u = x/4
du/dx = 1/4
...
y=ln(u)
dy/du = 1/u
1/u = 4/x

therefore dy/du *du/dx = dy/dx (in this case behaves like a fraction and du cancels)
so 4/x * 1/4 = 1/x
dy/dx = 1/x
I know that's works but I would do this:

Therefore

The ln(4) is a constant so it becomes 0 after differentiating
7. (Original post by asinghj)
I know that's works but I would do this:

Therefore

The ln(4) is a constant so it becomes 0 after differentiating
Oh I see what you mean. Yes! that is a much nicer way of doing it for this question!
8. (Original post by Kawaii289)
Oh I see what you mean. Yes! that is a much nicer way of doing it for this question!
I used to do it the way you did it, but my teacher showed this way and it becomes so much simpler

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