What is the derivative of #[ln(2-3x)/lnx]^[arccos(2-5x)]#?

Question

Evelyn Agard · Accepted Answer

#y^' = [ln(2-3)/lnx]^(arccos(2-5x)) * [5/sqrt(1-(2-5x)^2) * ln(ln(2-3x)/lnx) + 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x)]#

!! EXTREMELY LONG ANSWER !!

The most important derivation rule you'll need to use to differentiate this function is the power rule for a variable base and a variable power.

#color(blue)(d/dx(f(x)^g(x)) = f(x)^g(x) * d/dx[ln(f(x)) * g(x)])#

In your case, you have #f(x) = ln(2-3x)/ln(x)# and #g(x) = arccos(2-5x)#. Now, I will assume that you know the derivative of #arccosx#

#d/dx(arccosx) = -1/(sqrt(1-x^2)#

Ok, buckle up because the calculations will be #color(red)("horrendous")#.

So, start your calculation by writing the derivative of #y = f(x)^g(x)# using the power rule. To keep the calculations as compact as possible, I'll use #f(x) =f# and #g(x) = g# from this point on.

#d/dx(y) = f^g * d/dx(ln(f) * g)" "color(purple)((1))#

Use the product rule to find

#d/dx(ln(f) * g) = [d/dxln(f)] * g + ln(f) * d/dx(g)" "color(purple)((2))#

Now break this calculation into two different ones. The first one will be

#d/dxln(f) = d/dx[ln(ln(2-3x)/lnx)]#

Use the quotient rule once and the chain rule twice, once for #ln(u)#, with #u = ln(2-3x)/lnx#, and once more for #ln(v)#, with #v = 2-3x)#.

#d/dx(ln(u)) = d/(du)ln(u) * d/dx(u) " "color(purple)((3))#

#d/dx(u) = ([d/dxln(2-3x)] * lnx - ln(2-3x) * d/dx(lnx))/(lnx)^2 " "color(purple)((4))#

Use the second chain rule substitution to get

#d/dx(lnv) = d/(dv)lnv * d/dx(v)#

#d/dx(lnv) = 1/v * d/dx(2-3x)#

#d/dx(ln(2-3x)) = 1/(2-3x) * (-3)#

Plug this back into #color(purple)((4))# to get

#d/dx(u) = (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^2x#

Now take this result back to #color(purple)((3))# to get

#d/dx(ln(u)) = 1/u * (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^2x#

#d/dx(ln(f)) = color(red)(cancel(color(black)(lnx)))/ln(2-3x) * (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^color(red)(cancel(color(black)(2)))x#

#d/dx(lnf) = 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx))#

From #color(purple)((2))#, focus on finding #d/dx(g)#

#d/dx(g) = d/dx(arccos(2-5x))#

You're going to have to use th chain rule once for #arccost#, with #t = 2-5x)#

#d/dx(arccost) = d/(dt)arccost * d/dx(t)#

#d/dxarccost = -1/(sqrt(1-t^2)) * d/dx(2-5x)#

#d/dx(arccos(2-5x)) = -1/sqrt(1-(2-5x)^2) * (-5)#

#d/dx(g) = 5/sqrt(1-(2-5x)^2)#

Take everyting back to #color(purple)((2))# to get

#d/dx(ln(f) * g) = 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x) + ln(ln(2-3x)/lnx) * 5/sqrt(1-(2-5x)^2)#

Finally, take this back to #color(purple)((1))# to get

#y^' = color(green)([ln(2-3)/lnx]^(arccos(2-5x)) * [5/sqrt(1-(2-5x)^2) * ln(ln(2-3x)/lnx) + 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x)])#

Emily Ammerman · Answer

undefined The derivative of $ \left[\frac{\ln(2-3x)}{\ln x}\right]^{\arccos(2-5x)} $ can be found using the chain rule and logarithmic differentiation.