Question

What is the derivative of $-\cos \left( x \right).\ln \left( \sec \left( x \right)+\tan \left( x \right) \right)$?

Answer 1

Hint: We need to find the derivative of the function given. Since it is not a general function but a product of two functions with composition involved, we will use several properties of derivatives such as the chain rule and the product rule. We will first use product rule and then for the second function which is the composition of logarithmic and secant tangent function, we will use chain rule.

Complete step-by-step answer:
We are given the product of two functions. We will apply product rule, which states that for any two continuous function $f\left(x\right)$ and $g\left(x\right)$, the derivative of their product is as follows:
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + \dfrac{d}{{dx}}f\left( x \right)g\left( x \right)$
Before putting the values of $f\left(x\right)$ and $g\left(x\right)$ in this, we will first find the derivatives of both the functions so that it becomes easier for us to substitute values afterwards.
We have:
$f\left(x\right)=-cos\left(x\right)$
So, we have:
$\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( -\cos \left( x \right) \right)=\dfrac{-d}{dx}\left( \cos \left( x \right) \right)=-\left( -\sin \left( x \right) \right)=\sin \left( x \right)$
Now, we find the derivative of $g\left(x\right)$ where:
$g\left(x\right)=\ln \left( \sec \left( x \right)+\tan \left( x \right) \right)$
We can also say that
$g\left( x \right)=\ln \left( h\left( x \right) \right)$
\[h\left( x \right)=\left( \sec \left( x \right)+\tan \left( x \right) \right)\]
So, we apply chain rule now:
$\dfrac{d}{dx}\left( g\left( x \right) \right)=\dfrac{dg}{dh}\times \dfrac{dh}{dx}=\dfrac{1}{h}\times \left( \sec \left( x \right)\tan \left( x \right)+{{\sec }^{2}}\left( x \right) \right)$
Plugging the value of $h\left(x\right)$ in this we obtain:
$\dfrac{d}{dx}\left( g\left( x \right) \right)=\dfrac{1}{\sec \left( x \right)+\tan \left( x \right)}\times \left( \sec \left( x \right)\left( \tan \left( x \right)+\sec \left( x \right) \right) \right)=\sec \left( x \right)$
Now, we put these in the product rule and we obtain the following:
$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=-\cos \left( x \right)\sec \left( x \right)+\sin \left( x \right)\ln \left( \sec \left( x \right)+\tan \left( x \right) \right)$
Now, since
$cos\left(x\right)\times sec\left(x\right)=1$
We get:
$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=-1+\sin \left( x \right)\ln \left( \sec \left( x \right)+\tan \left( x \right) \right)$
Hence, we have found the derivative of the function provided.

Note: Do not mix product rule and chain rule. While applying the formula for chain rule, do not forget to differentiate till the last function. Most often, we forget to add a negative sign in the expression as well, which leads to an incorrect result. So, be very careful while calculating the derivative of any complex function involving composition or multiplication.