Question

Differentiate and simplify please help?

Answer 1

`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)`

Answer 2

`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)`