Diffrentiation(logs) cannot solve please help (1 Viewer)

[Steven12]

can someone out there explain to me why
when you diffrentiate

Y=e To power of Ln(x) equals to exactly 1

and how do you f'(x) y=Ln(x+1)/(1-y)

and how to f'(x) y=Ln(x+3)(x-1)


Iam not really bright at maths so someone out there who knows please help

 

[CM_Tutor]

y = e^{ln x} differentiates to dy/dx = 1 as e^{ln x} = x.

For the others, you can differentiate them as they stand using chain and quotient rules for the first and chain and product rules for the second. However, this is the hard way.

The trick with differentiating anything complicated with a log in it is to apply log laws first.

So, for y = ln [(x + 1) / (1 - x)] - which is what I assume you meant - use log law log(a / b) = log a - log b
y = log (x + 1) - log(1 - x). Now we can differentiate (remember, if f(x) = ln (Ax + B), f'(x) = A / (Ax + B)
dy/dx = [1 / (x + 1)] - [-1 / (1 - x)]
= [1 / (x + 1)] + [1 / (1 - x)]
= [(1 - x) + (x + 1)] / (x + 1)(1 - x)
= (1 - x + x + 1) / (1² - x²)
= 2 / (1 - x²)

And, for y = ln [(x + 3)(x - 1)], use log law log(ab) = log a + log b
y = log (x + 3) + log(x - 1). Now we can differentiate (remember, if f(x) = ln (Ax + B), f'(x) = A / (Ax + B)
dy/dx = [1 / (x + 3)] + [1 / (x - 1)]
= [(x - 1) + (x + 3)] / (x + 3)(x - 1)
= (x - 1 + x + 3) / (x² + 2x - 3)
= (2x + 2) / (x² + 2x - 3)

 

[Steven12]

ohe my god

far out man. i should have known


thanks man. appreciated it