Differentiate and simplify please help?

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2 Answers
Feb 13, 2018

#x^(tanx)(lnxsec^2x+1/xtanx)#

Explanation:

Express #x^tanx# as power of e:

#x^tanx=e^ln(x^tanx)=e^(lnxtanx)#

#=d/dxe^(lnxtanx)#

Using the chain rule, #d/dxe^(lnxtanx)=(de^u)/(du)((du)/dx),# where #u=lnxtanx# and #d/(du)(e^u)=e^u#

#=(d/dx(lnxtanx))e^(lnxtanx)#

Express #e^(lnxtanx)# as a power of x:
#e^(lnxtanx)=e^ln(x^tanx)=x^tanx#

#=x^tanx. d/(dx)(lnxtanx)#

Use the product rule, #d/(dx)(uv)=v(du)/(dx)+u(dv)/(dx)#
, where #u=lnx# and #v=tanx#

#=lnx d/(dx)(tanx)+d/(dx)(lnxtanx)x^tanx#

The derivate of #tanx# is #sec^2x#

#=x^tanx(sec^2xlnx+(d/(dx)(lnx))tanx)#

The derivative of #lnx# is #1/x#

#=x^tanx(lnxsec^2x+1/xtanx)#

Feb 13, 2018

#dy/dx=(sec^2(x)ln(x)+tan(x)/x)x^tan(x)#

Explanation:

We shall use logarithmic differentiation - that is, we will take the natural log of both sides and differentiate implicitly w.r.t #x#

Given: #y=x^tan(x)#

Take the natural log (#ln#) of both sides:

#ln(y)=ln(x^tan(x))#

Applying the power rule of natural log #ln(a)^b=b*ln(a)#

#ln(y)=tan(x)*ln(x)#

Differentiate both sides implicitly w.r.t #x#

#1/y*dy/dx=color(blue)(sec^2(x)ln(x)+tan(x)/x)# (See work below)

To differentiate the RHS, we will need to use the product rule!

We have #d/dx[tan(x)*ln(x)]#

Let #f(x)=tan(x)# and #g(x)=ln(x)#

Thus, #f'(x)=sec^2(x)# and #g'(x)=1/x#

By the product rule: #d/dx[f(x)*g(x)]=f'(x)g(x)+f(x)g(x)#

Substituting we get:

#d/dx[tan(x)*ln(x)]=sec^2(x)*ln(x)+tan(x)*1/x#

Simplifying...

#d/dx[tan(x)*ln(x)]=sec^2(x)*ln(x)+tan(x)/x#

Going back to what we had before:

#1/y*dy/dx=sec^2(x)ln(x)+tan(x)/x#

We want to isolate #dy/dx# so we multiply both sides by #y#

#cancelcolor(red)y*1/cancely*dy/dx=(sec^2(x)ln(x)+tan(x)/x)*color(red)y#

#dy/dx=(sec^2(x)ln(x)+tan(x)/x)*color(red)y#

We want to write everything in terms of #x# but we have this #color(red)y# in the way. You may recall that #color(red)y# is given to us in the very beginning. #color(red)(y=x^tan(x))#

#:.dy/dx=(sec^2(x)ln(x)+tan(x)/x)*x^tan(x)#