How do you differentiate #y = (cos 7x)^x#?

1 Answer
Feb 20, 2018

#dy/dx = (cos(7x))^x*(ln(cos(7x))-7x(tan(7x)))#

Explanation:

This is nasty.

#y = (cos(7x))^x#

Start by taking the natural logarithm of either side, and bring the exponent #x# down to be the coefficient of the right hand side:

#rArr lny = xln(cos(7x))#

Now differentiate each side with respect to #x#, using the product rule on the right hand side. Remember the rule of implicit differentiation: #d/dx(f(y)) = f'(y)*dy/dx#

#:.1/y*dy/dx = d/dx(x)*ln(cos(7x)) + d/dx(ln(cos(7x)))*x#

Using the chain rule for natural logarithm functions - #d/dx(ln(f(x))) = (f'(x))/f(x)# - we can differentiate the #ln(cos(7x))#

#d/dx(ln(cos(7x))) = -7sin(7x)/cos(7x) = -7tan(7x)#

Returning to the original equation:

#1/y*dy/dx = ln(cos(7x)) - 7xtan(7x)#

Now we can substitute the original #y# as a function of #x# value from the start back in, so as to remove the errant #y# on the left hand side. Multiplying both sides by #y#:

#dy/dx = (cos(7x))^x*(ln(cos(7x))-7x(tan(7x)))#