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)