Question

Find the derivative of ${\sin ^2}x$ using the first principles?

Answer 1

Hint: We have to find the derivative of ${\sin ^2}x$ using first principle. First principle is that if there is a function $f\left( x \right)$ then the derivative of $f\left( x \right)$ will be
$ \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
We have explained this principle in detail below.

Complete step-by-step answer:
In this question, we need to find the derivative of ${\sin ^2}x$ using the first principles.
Now, we could easily differentiate ${\sin ^2}x$ using the usual differentiation method, but we are going to see the long method, that is the first principle in this question.
Now, first of all let us understand what first principle is.
Let $f\left( x \right)$ be a function in its domain. A function defined such that if $\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}} \right]$ exists, it is said to be the derivative of $f\left( x \right)$. This is known as the first principle of derivative and it is also known as the Delta Method.
Now, here our $f\left( x \right)$ is ${\sin ^2}x$.
$ \to f\left( x \right) = {\sin ^2}x$
And the derivative of $f\left( x \right)$ will be equal to
$ \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
$ \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^2}\left( {x + h} \right) - {{\sin }^2}x}}{h}$
Now, we know that $\sin \left( {x + h} \right) = \sin x\cosh + \cos x\sinh $. Therefore,
$ \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {\sin x\cosh + \cos x\sinh } \right)}^2} - {{\sin }^2}x}}{h}$
Now, using the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get
\[
   \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^2}x{{\cos }^2}h + {{\cos }^2}x{{\sin }^2}h + 2\sin x\cosh \sinh \cos x - {{\sin }^2}x}}{h} \\
   \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^2}x\left( {{{\cos }^2}h - 1} \right) + 2\sin x\cosh \sinh \cos x + {{\cos }^2}x{{\sin }^2}h}}{h} \\
 \]
Now, e know the identity ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, ${\cos ^2}h - 1 = - {\sin ^2}h$. Therefore, substituting it in above equation, we get
\[
   \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - {{\sin }^2}x{{\sin }^2}h + 2\sin x\cosh \sinh \cos x + {{\cos }^2}x{{\sin }^2}h}}{h} \\
   \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^2}h\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + 2\sin x\cosh \sinh \cos x}}{h} \\
   \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sinh }}{h} \cdot \sinh \left( {{{\cos }^2}x - {{\sin }^2}x} \right) + 2\sin x\cosh \cos x \cdot \dfrac{{\sinh }}{h}} \right] \\
 \]
Now, using the fundamental trigonometric calculus limits,
$ \to \mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin \theta }}{\theta } = 1$
Therefore, we get
\[ \to f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left[ {\left( 1 \right) \cdot \sinh \left( {{{\cos }^2}x - {{\sin }^2}x} \right) + 2\sin x\cosh \cos x \cdot \left( 1 \right)} \right]\]
Now, put the value of h, we get
\[
   \to f'\left( x \right) = \left[ {\left( 1 \right) \cdot \sin 0\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + 2\sin x\cos 0\cos x \cdot \left( 1 \right)} \right] \\
   \to f'\left( x \right) = \left[ {\left( 1 \right) \cdot \left( 0 \right)\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + 2\sin x\left( 1 \right)\cos x\left( 1 \right)} \right] \\
   \to f'\left( x \right) = \left[ {0 + 2\sin x\cos x} \right] \\
   \to f'\left( x \right) = 2\sin x\cos x \\
 \]
Hence, we have calculated the derivative of ${\sin ^2}x$ using first principle.

Note: We can also find the derivative of ${\sin ^2}x$ normally using the chain rule.
$ \to f\left( x \right) = {\sin ^2}x$
Here, we have to differentiate both $\sin x$ and its power.
$ \to f'\left( x \right) = 2\sin x\cos x$