Question

What is the derivative of $\sec \left( {x - {x^2}} \right)$ ?

Answer 1

Generally in Mathematics, derivative refers to the rate of change of a function with respect to a variable. Here, we are applying the chain rule to find the required answer. We need to use the chain rule in three steps.
First, we need to identify the inner function and replace it by $u$ and then rewrite the outer function with respect to the inner function.
The second step is to find the derivatives of both functions.
The third step is to substitute the given expression and the derivatives in the variable $u$ and then simplify it. This is nothing but the chain rule.
The formula for the chain rule for differentiation is as follows:
If $y = f(x)$ , then
$f'(x) = \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Also, $\dfrac{d}{{dx}}\sec x = \sec x\tan x$

Complete step-by-step answer:
Consider $y = \sec (x - {x^2})$ ..(1)
Then let $u = (x - {x^2})$
Now, we have to differentiate the above equation with respect to x, we get,
$\dfrac{{du}}{{dx}} = 1 - 2x$
Now, equation (1) becomes
$y = \sec (u)$
Now, we have to differentiate the above equation with respect to u, we get,
$\dfrac{{dy}}{{dx}} = \sec (u) \times \tan (u)$
Now, we need to use the chain rule of differentiation formula,
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dx}} \times \dfrac{{du}}{{dx}}$
Applying the obtained result in the above formula,
$\dfrac{{dy}}{{dx}} = (\sec (u) \times \tan (u))(1 - 2x)$
Now, substitute $u = x - {x^2}$ in the above equation,
$\dfrac{{dy}}{{dx}} = sex(x - {x^2})\tan (x - {x^2})(1 - 2x)$
Hence, the derivative of $\sec (x - {x^2}) = (1 - 2x)sec(x - {x^2})\tan (x - {x^2})$

Note: We can also find our required solution without using the chain rule.
Apply $\sec x = \dfrac{1}{{\cos x}}$ and then apply the quotient rule, to get the desired solution
The formula for the quotient rule is, $\dfrac{u}{v} = \dfrac{{u'v - v'u}}{{{v^2}}}$ .
We know that the derivative of $\cos x$ is $- \sin x$ and the derivative of 1 is 0.
Then, $\dfrac{d}{{dx}}(\sec (x) = \dfrac{{(\cos x)(0) - (1)( - \sin x)}}{{{{(\cos x)}^2}}}$
$= \dfrac{{\sin x}}{{\cos x \times \cos x}}$ $$
$= \dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\cos x}} = \sec x\tan x$
Now we are putting $x - {x^2}$ in the place of x, we get,
$\dfrac{d}{{dx}}\sec (x - {x^2})\tan (x - {x^2})\dfrac{d}{{dx}}(x - {x^2})$
The derivative of $(x - {x^2})$ is $1 - 2x$ so that we can obtain our desired solution,
$\dfrac{d}{{dx}}\sec (x - {x^2}) = (1 - 2x)\sec (x - {x^2})tan(x - {x^2})$ .