Question

Find the derivative of $\sqrt[3]{\sin x}$ from first principles.

Answer 1

To answer this question, we need to use the first principle of differentiation which is given by, $\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.$ We use this to calculate the derivative of $\sqrt[3]{\sin x}$ . While calculating, we need to use L'Hospital's rule because we get a $\dfrac{0}{0}$ form while solving. Then we substitute the value of the limit to obtain the final answer.

Complete step by step solution:
Let us consider the given question as the function $f\left( x \right),$
$\Rightarrow f\left( x \right)=\sqrt[3]{\sin x}$
We simplify this by using the principles of differentiation. This is given by the formula $\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.$ We now substitute the value of the function $f\left( x \right)$ in the above equation.
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\displaystyle \lim_{h \to 0}\dfrac{\sqrt[3]{\sin \left( x+h \right)}-\sqrt[3]{\sin x}}{h}$
Substituting the value of the limit as h as 0,
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\dfrac{\sqrt[3]{\sin \left( x+0 \right)}-\sqrt[3]{\sin x}}{0}$
Subtracting the same two terms in the numerator gives us a 0.
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\dfrac{\sqrt[3]{\sin x}-\sqrt[3]{\sin x}}{0}$
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\dfrac{0}{0}$
This is an indeterminate form and can be simplified using the L Hospital’s rule which is nothing but the application of partial differentiation of the numerator and denominator separately.
Differentiating the numerator with respect to h,
$\Rightarrow \dfrac{d}{dh}\left( \sqrt[3]{\sin \left( x+h \right)}-\sqrt[3]{\sin x} \right)$
We use the differentiation of the composite function method and differentiate the outside function first and multiply it with the inside function. We know the differentiation of ${{x}^{\dfrac{1}{3}}}$ is given as,
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{\dfrac{1}{3}}} \right)=\dfrac{1}{3}{{x}^{\dfrac{1}{3}-1}}$
Subtracting the powers and simplifying,
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{\dfrac{1}{3}}} \right)=\dfrac{1}{3}{{x}^{-\dfrac{2}{3}}}$
This is the outside function. Here $x$ is $\sin \left( x+h \right).$ $\sqrt[3]{\sin x}$ term is a constant since it does not contain h and we know the differentiation of a constant is 0. The differentiation of the inside function is $\sin \left( x+h \right),$ and we know the differentiation of $\sin x$ is $\cos x.$ Multiplying the two,
$\Rightarrow \dfrac{1}{3}\sin {{\left( x+h \right)}^{-\dfrac{2}{3}}}.\cos \left( x+h \right)$
Now, we differentiate the denominator h and we know,
$\Rightarrow \dfrac{dh}{dh}=1$
Substituting back in the equation after applying L Hospital’s rule,
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{1}{3}\sin {{\left( x+h \right)}^{-\dfrac{2}{3}}}.\cos \left( x+h \right)}{1}$
Now applying the limit,
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\dfrac{\dfrac{1}{3}\sin {{\left( x+0 \right)}^{-\dfrac{2}{3}}}.\cos \left( x+0 \right)}{1}$
Simplifying,
$\Rightarrow \dfrac{d}{dx}\sqrt[3]{\sin x}=\dfrac{1}{3}\sin {{\left( x \right)}^{-\dfrac{2}{3}}}.\cos \left( x \right)$

Hence, we have used the first principle of differentiation to calculate the derivative of $\sqrt[3]{\sin x}$ which is $\dfrac{1}{3}\sin {{\left( x \right)}^{-\dfrac{2}{3}}}.\cos \left( x \right).$

Note: Students need to know the basic differentiation formulae and need to know the first principle method of differentiation which is given by the important formula, $\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}.$ They need to know the concept of limits too and the application of L'Hospital's rule to simplify this question easily.