Take the natural logarithm of both sides.
#ln[y] = ln[(lnx)^(tanx)]#
Simplify the right-hand side using the rule #ln[a^n} = nlna#:
#ln[y] = tanx[ln(lnx)]#
Differentiate both sides.
#1/y(dy/dx) = square#
Inset: #square#
The derivative of the right hand side is fairly complex. We have to first find the derivative of #ln(lnx)# using the chain rule, followed by the derivative of #tanx xx ln(lnx)#, using the product rule.
Let #y = ln(u)# and #u = lnx#. Then #y' = 1/u# and #u' = 1/x#.
#u' = 1/u xx 1/x = 1/(lnx) xx 1/x = 1/(xlnx)#
Now we have to find the derivative for #tanx# so that we have enough information to apply the product rule.
#tanx = sinx/cosx#
By the quotient rule:
#(tanx)' = (cosx xx cosx - (-sinx xx sinx))/(cosx)^2#
#(tanx)' = (cos^2x + sin^2x)/(cos^2x)#
Applying the pythagorean identity #cos^2x + sin^2x = 1#:
#(tanx)' = 1/(cos^2x) = sec^2x#
Next, apply the product rule:
Let #y = g(x) xx h(x)#, where #g(x) = ln(lnx) and h(x) = tanx#:
#y' = (1/(xlnx) xx tanx + ln(lnx) xx sec^2x)#
#y' = tanx/(xlnx) + sec^2xln(lnx)#
Putting this into the place of the #square#:
#1/y(dy/dx) = tanx/(xlnx) + sec^2xln(lnx)#
#dy/dx = ( tanx/(xlnx) + sec^2xln(lnx))/(1/y)#
#dy/dx = y( tanx/(xlnx) + sec^2xln(lnx))#
#dy/dx= (lnx)^(tanx)( tanx/(xlnx) + sec^2xln(lnx))#
I know it's quite messy, but it worked!!
Practice exercises:
- Determine the derivative of the given relation.
a) #f(x) = log_2(sin(3x + 4))#
b) #g(x) = (2lnx)^(sec(3x + 5))#
Hopefully this helps, and good luck!
