Question

How do you find the derivative of $\cos (5x)$?

Answer 1

Hint: The given trigonometry is $\cos (5x)$
 Differentiate using the chain rule, which states that
$\dfrac{d}{{dx}}[f(g(x))]$ is $f'(g(x))g'(x)$
To apply the chain rule, set $u$ as $5x$ and get the required result.

Complete step-by-step solution:
The given trigonometry is $\cos (5x)$
Let us consider $y = \cos (5x)$
Differentiate using the chain rule, which states that
$\dfrac{d}{{dx}}[f(g(x))]$ is $f'(g(x))g'(x)$ where
$f(x) = \cos (x)$and$g(x) = 5x$
To apply the chain rule, set $u$ as $5x$, hence we get
$\Rightarrow$$\dfrac{d}{{dx}}[\cos (u)]\dfrac{d}{{dx}}[5x]$
The derivation of $\cos (u)$ with respect to $u$ is $ - \sin (u)$
$\Rightarrow$$ - \sin (u)\dfrac{d}{{dx}}[5x]$
Replace all occurrences of $u$ with$5x$
$\Rightarrow$$ - \sin (5x)\dfrac{d}{{dx}}[5x]$
Differentiate
Since $5$ is constant with respect to $x$, the derivative of $5x$ with respect to $x$ is $5\dfrac{d}{{dx}}[x]$
$\Rightarrow$$ - \sin (5x)\left( {5\dfrac{d}{{dx}}[x]} \right)$
Multiply $5$ by $ - 1$
$\Rightarrow$$ - 5\sin (5x)\dfrac{d}{{dx}}[x]$
Differentiate use the power rule which states that $\dfrac{d}{{dx}}[{x^n}]$ is $n{x^{n - 1}}$ where $n = 1$
$\Rightarrow$$ - 5\sin (5x) \cdot 1$
Multiply $ - 5$ by $1$
$\Rightarrow$$ - 5\sin (5x)$

Then the derivative of $\cos (5x)$ is $ - 5\sin (5x)$.

Note: Suppose that we have two functions $f(x)$ and $g(x)$ they are both differentiable.
If we define $F(x) = (f \circ g)(x)$ then the derivative of $F(x)$ is,
$F'(x) = f'(g(x))g'(x)$
If we have $y = f(u)$ and $u = g(x)$ then the derivative of $y$ is
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}\dfrac{{du}}{{dx}}$
Each of these forms has its uses, however we will work mostly with the first form in this class. To see the proof chain rule see the proof of various derivative formulas. We have taken a lot of derivatives. However, if you look back they have all been functions similar to the following kinds of functions.
$R(z) = \sqrt z $ $f(t) = {t^{30}}$ $y = \tan (x)$
These are all fairly simple functions in that wherever the variable appears it is by itself. None of our rules will work on these functions and yet some of these functions are closer to the derivatives that we are liable to run into than the functions in the first set. Let’s take the first one for example. The definition of the compute this derivative.