Question

How do you find the derivative of $y = \dfrac{4}{{\cos x}}$?

Answer 1

Hint: We have to find $y'$. For this differentiate $y$ with respect to $x$. Then, use the property that the differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function. Then, use trigonometry identity $\cos \theta \times \sec \theta = 1$ to replace $\dfrac{1}{{\cos x}}$ with $\sec x$. Then, use the property that the differentiation of secant function is a product of secant and tangent function and get the desired result.

Formula used: The differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
The differentiation of secant function is a product of secant and tangent function.
i.e., $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
Trigonometric identity: $\cos \theta \times \sec \theta = 1$

Complete step-by-step solution:
We have to find the derivative of $y = \dfrac{4}{{\cos x}}$.............…(i)
So, differentiate $y$ with respect to $x$.
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{4}{{\cos x}}} \right)$...........…(ii)
Now, using the property that the differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
So, in above differentiation (ii), constant $4$ can be taken outside the differentiation.
$ \Rightarrow \dfrac{{dy}}{{dx}} = 4\dfrac{d}{{dx}}\left( {\dfrac{1}{{\cos x}}} \right)$...........…(iii)
Now, use trigonometry identity $\cos \theta \times \sec \theta = 1$ to replace $\dfrac{1}{{\cos x}}$ with $\sec x$.
So, differentiation (iii) can be written as
$ \Rightarrow \dfrac{{dy}}{{dx}} = 4\dfrac{d}{{dx}}\left( {\sec x} \right)$............…(iv)
Now, using the property that the differentiation of secant function is a product of secant and tangent function.
i.e., $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
So, in above differentiation (iv), we can use above property and find the derivative of $y$.$ \Rightarrow \dfrac{{dy}}{{dx}} = 4\sec \left( x \right)\tan \left( x \right)$

Therefore, the derivative is $y' = 4\sec \left( x \right)\tan \left( x \right)$.

Note: We can also find the derivative using the limit definition of derivative.
Definition of the Derivative:
The derivative of $f\left( x \right)$ with respect to $x$ is the function $f'\left( x \right)$ and is defined as,
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$..........…(1)
So, all we really need to do is to plug this function into the definition of the derivative, (1), and do some algebra.
First plug the function into the definition of the derivative.
 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
$ \Rightarrow y'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{4}{{\cos \left( {x + h} \right)}} - \dfrac{4}{{\cos \left( x \right)}}}}{h}$
Now, we know that we can’t just plug in $h = 0$ since this will give us a division by zero error. So, we are going to have to do some work.
$ \Rightarrow y'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{\cos \left( {x + h} \right)}} - \dfrac{1}{{\cos \left( x \right)}}}}{h}$
$ \Rightarrow y'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( x \right) - \cos \left( {x + h} \right)}}{{h\cos \left( x \right)\cos \left( {x + h} \right)}}$
As $\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right)$.
$ \Rightarrow y'\left( x \right) = 4\mathop {\lim }\limits_{h \to 0} \dfrac{{2\sin \left( {\dfrac{{2x + h}}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{{h\cos \left( x \right)\cos \left( {x + h} \right)}}$
$ \Rightarrow y'\left( x \right) = 4\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {\dfrac{{2x + h}}{2}} \right)}}{{h\cos \left( x \right)\cos \left( {x + h} \right)}} \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {h/2} \right)}}{{h/2}}} \right]$
As $\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {h/2} \right)}}{{h/2}} = 1$.
$ \Rightarrow y'\left( x \right) = 4 \times \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)\sin \left( x \right)}} \times 1$
$ \Rightarrow y'\left( x \right) = 4\sec \left( x \right)\tan \left( x \right)$
Therefore, the derivative is $y' = 4\sec \left( x \right)\tan \left( x \right)$.