Question

The derivative of $\sin x$ is equal to $\cos x$ , $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$ and $\dfrac{d}{{dx}}\left[ {\sin f\left( x \right)} \right] = \cos f\left( x \right) \times \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$ . Use the above information to find the derivatives of the following.
a. $\sin {x^2}$
b. $\sin {x^a}$
c. $\sin \left( {\log x} \right)$
d. ${\sin ^{25}}\left( {{x^a}} \right)$
e. $\sin \left( {\sin {x^7}} \right)$ f. $\sin \left( {ax + b} \right)$

Answer 1

Hint : The derivative of sine is cosine. But when the sine is applied to another function (can be exponential, logarithmic etc), the derivative will be different. Including the cosine of function, derivative of the function must also be multiplied. Derivatives of Exponential function, logarithmic function, linear function are different. Use this info and below formulas to answer the given question.
Formulas used:
1. $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
2. $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}$
3. $\dfrac{d}{{dx}}{\left[ {f\left( x \right)} \right]^n} = n.{\left[ {f\left( x \right)} \right]^{n - 1}} \times \dfrac{d}{{dx}}f\left( x \right)$
4. $\dfrac{d}{{dx}}\left( {ax + b} \right) = a$

Complete step by step solution:
We are given that $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$ and $\dfrac{d}{{dx}}\left[ {\sin f\left( x \right)} \right] = \cos f\left( x \right) \times \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$ .
Using this we have to find the derivatives of sine functions given.
a. $\sin {x^2}$
On comparing the given sine function with $\sin f\left( x \right)$ , we get $f\left( x \right) = {x^2}$
Therefore, $\dfrac{d}{{dx}}\left( {\sin {x^2}} \right) = \cos \left( {{x^2}} \right) \times \dfrac{d}{{dx}}\left( {{x^2}} \right)$
We know that $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}$ , here n = 2.
$\Rightarrow \dfrac{d}{{dx}}\left( {\sin {x^2}} \right) = \cos \left( {{x^2}} \right) \times \left( {2 \times {x^{2 - 1}}} \right) \\\ \Rightarrow \dfrac{d}{{dx}}\left( {\sin {x^2}} \right) = \cos \left( {{x^2}} \right) \times 2x \\\ \therefore \dfrac{d}{{dx}}\left( {\sin {x^2}} \right) = 2x.\cos {x^2} \\\$
b. $\sin {x^a}$
On comparing the given sine function with $\sin f\left( x \right)$ , we get $f\left( x \right) = {x^a}$
Therefore, $\dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = \cos \left( {{x^a}} \right) \times \dfrac{d}{{dx}}\left( {{x^a}} \right)$
We know that $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}$ , here n = a.
$\Rightarrow \dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = \cos \left( {{x^a}} \right) \times \left( {a \times {x^{a - 1}}} \right) \\\ \Rightarrow \dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = \cos \left( {{x^a}} \right) \times a.{x^{a - 1}} \\\ \therefore \dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = a.{x^{a - 1}}.\cos {x^a} \\\$
c. $\sin \left( {\log x} \right)$
On comparing the given sine function with $\sin f\left( x \right)$ , we get $f\left( x \right) = \log x$
Therefore, $\dfrac{d}{{dx}}\left[ {\sin \left( {\log x} \right)} \right] = \cos \left( {\log x} \right) \times \dfrac{d}{{dx}}\left( {\log x} \right)$
We know that $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$ , so we are substituting this above.
$\dfrac{d}{{dx}}\left[ {\sin \left( {\log x} \right)} \right] = \cos \left( {\log x} \right) \times \dfrac{1}{x} \\\ \therefore \dfrac{d}{{dx}}\left[ {\sin \left( {\log x} \right)} \right] = \dfrac{1}{x}.\cos \left( {\log x} \right) \\\$
d. ${\sin ^{25}}\left( {{x^a}} \right)$
We can write the given sine function also as ${\left( {\sin {x^a}} \right)^{25}}$ . It is in the form of ${\left[ {f\left( x \right)} \right]^n}{\text{ where f}}\left( x \right) = \sin {x^a},n = 25$
We know that $\dfrac{d}{{dx}}{\left[ {f\left( x \right)} \right]^n} = n.{\left[ {f\left( x \right)} \right]^{n - 1}} \times \dfrac{d}{{dx}}f\left( x \right)$
$\dfrac{d}{{dx}}\left[ {{{\left( {\sin {x^a}} \right)}^{25}}} \right] = 25 \times {\left( {\sin {x^a}} \right)^{25 - 1}} \times \dfrac{d}{{dx}}\left( {\sin {x^a}} \right)$
From the above second solution, we got $\dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = a.{x^{a - 1}}.\cos {x^a}$
$\dfrac{d}{{dx}}\left[ {{{\left( {\sin {x^a}} \right)}^{25}}} \right] = 25 \times {\left( {\sin {x^a}} \right)^{24}} \times a \times {x^{a - 1}} \times \cos {x^a} \\\ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {\sin {x^a}} \right)}^{25}}} \right] = 25a.{x^{a - 1}}.{\left( {\sin {x^a}} \right)^{24}}\cos {x^a} \\\$
e. $\sin \left( {\sin {x^7}} \right)$
On comparing the given sine function with $\sin f\left( x \right)$ , we get $f\left( x \right) = \sin {x^7}$
Therefore, $\dfrac{d}{{dx}}\left[ {\sin \left( {\sin {x^7}} \right)} \right] = \cos \left( {\sin {x^7}} \right) \times \dfrac{d}{{dx}}\left( {\sin {x^7}} \right)$
From the above second solution, we got $\dfrac{d}{{dx}}\left( {\sin {x^a}} \right) = a.{x^{a - 1}}.\cos {x^a}$ , here a = 7
$\dfrac{d}{{dx}}\left[ {\sin \left( {\sin {x^7}} \right)} \right] = \cos \left( {\sin {x^7}} \right) \times 7 \times {x^{7 - 1}} \times \cos {x^7} \\\ \Rightarrow \dfrac{d}{{dx}}\left[ {\sin \left( {\sin {x^7}} \right)} \right] = 7{x^6}.\cos \left( {\sin {x^7}} \right)\cos {x^7} \\\$
f. $\sin \left( {ax + b} \right)$
On comparing the given sine function with $\sin f\left( x \right)$ , we get $f\left( x \right) = ax + b$
Therefore, $\dfrac{d}{{dx}}\left[ {\sin \left( {ax + b} \right)} \right] = \cos \left( {ax + b} \right) \times \dfrac{d}{{dx}}\left( {ax + b} \right)$
We know that $\dfrac{d}{{dx}}\left( {ax + b} \right) = a$
So on substituting the value, we get
$\dfrac{d}{{dx}}\left[ {\sin \left( {ax + b} \right)} \right] = \cos \left( {ax + b} \right) \times a \\\ \Rightarrow \dfrac{d}{{dx}}\left[ {\sin \left( {ax + b} \right)} \right] = a.\cos \left( {ax + b} \right) \;$

Note : Be careful with the derivatives of exponential functions, linear functions, logarithmic functions etc. and we might confuse ${x^a}$ with ${a^x}$ , so be cautious about this as their derivatives are completely different. Derivative of ${x^a}$ is $a.{x^{a - 1}}$ whereas derivative of ${a^x}$ is ${a^x}\log a$ . Do not swap the base with exponent as you can see the result might completely go wrong.