Question

Find the derivative of `(tanx)^secx + (secx)^cotx` W.R.T. x?

Answer 1

Derivative of `(tanx)^secx+(secx)^cotx` is

`sec^3x(tanx)^(secx-1)+(secx)^(cotx)`

Explanation:

We use here Chain Rule - In order to differentiate a function of a function, say `y, =f(g(x))`, where we have to find `(dy)/(dx)`, we need to do (a) substitute `u=g(x)`, which gives us `y=f(u)`. Then we need to use a formula called Chain Rule, which states that `(dy)/(dx)=(dy)/(du)xx(du)/(dx)`. In fact if we have something like `y=f(g(h(x)))`, we can have `(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)`.

Here let us split the function in two parts as follows:

`f(x)=(tanx)^secx` and `g(x)=(secx)^cotx`

Now we will find derivative of each and then add them up as

`d/(dx)(f(x)+g(x))=(df)/(dx)+(dg)/(dx)`

As `f(x)=(tanx)^secx`, we can write it as `f(x)=(u(x))^secx`, where `u(x)=tanx` and hence

`(df)/(du)=secx(u(x))^(secx-1)` and `(du)/(dx)=sec^2x`

Hence `(df)/(dx)=secx(u(x))^(secx-1)xxsec^2x`

= `secx(tanx)^(secx-1)sec^2x=sec^3x(tanx)^(secx-1)`

Similarly `(dg)/(dx)=cotx(secx)^(cotx-1)xxsecxtanx`

= `secx(secx)^(cotx-1)=(secx)^(cotx)`

Hence derivative of `(tanx)^secx+(secx)^cotx` is

`sec^3x(tanx)^(secx-1)+(secx)^(cotx)`

Answer 2

`(tanx)^secx(secxtanx(ln(tanx))+sec^3x/tanx)+(secx)^cotx(1-csc^2xln(secx))`

Explanation:

Both of these derivatives can be found using logarithmic differentiation.

Letting `w=(tanx)^secx`, we see that

`ln(w)=ln((tanx)^secx)=secxln(tanx)`

Differentiating, we see that

`1/w((dw)/dx)=secxtanx(ln(tanx))+secx(sec^2x/tanx)`

`(dw)/dx=(tanx)^secx(secxtanx(ln(tanx))+sec^3x/tanx)`

And a similar method is used for `v=(secx)^cotx`:

`ln(v)=ln((secx)^cotx)=cotxln(secx)`

`1/v((dv)/dx)=-csc^2xln(secx)+cotx((secxtanx)/secx)`

`(dv)/dx=(secx)^cotx(1-csc^2xln(secx))`

Thus, the derivative of `(tanx)^secx+(secx)^cotx` is `(tanx)^secx(secxtanx(ln(tanx))+sec^3x/tanx)+(secx)^cotx(1-csc^2xln(secx))`.