using logarithmic differentiation to evaluate f'(x) of f(x)=x^8cos^3x/sqrt(x-1)

Question

evaluate to find f' (x). I have some info on how to dissect it using ln, like =ln(x^8)+ln(cos^3x)-ln(sqrt(x-1))

Answer 1

Let y = x⁸cos³x/√(x-1)

 

lny = ln(x⁸) + ln(cos³x) - ln(x-1)^1/2

 

lny = 8lnx +3ln(cosx) - ½ln(x-1)

 

So, y'/y = 8/x - 3tanx - 1/(2x-2)

 

y' = y[8/x - 3tanx - 1/(2x-2)]

 

y' = [x⁸cos³x/√(x-1)][8/x - 3tanx - 1/(2x-2)]

 

 

Answer 2

We know that  d/dx(ln(x)) is 1 / x.

 

 

Use the chain rule to differentiate.  First, we want to compact the logarithm so that is a single logarithmic expression.

 

f(x) = ln[(x⁸cos³x) / (x - 1)^1/2]

 

 

Now, we can find the derivative of the function.  This will be a combination of chain, product and quotient rule.

 

f'(x) = [((8x⁷cos³x - 3x⁸cos²xsinx)(x - 1)^1/2 - (1/2)(x - 1)^-1/2(x⁸cos³x)) / (x - 1)^1/2] * [(x - 1)^1/2 / (x⁸cos³x)]

 

 

Now just simplify the derivative.

         

Answer 3

f(x) = f(x)=(x^8)(Cos^3x)/sqrt(x-1)   ⇐ I've introduced grouping symbols to make absolutely clear what's what.

 

I see this as g(x) / √(x-1) where g(x) = x⁸ (cos x)³ i.e., the original function's numerator; the denominator is that square root.

 

The quotient rule gives f'(x) = (u'v-uv')÷v² and in this instance u = g(x) and v = (x-1)^½.

u'(x)  = 8x⁷(cos x)³ +x⁸(-3cos²x)(sin x) and v' = ½(x-1)^-½

 

^{Now you just need to assemble all the parts, according to the rule for differentiation of (u/v), as stated above.}