Question

Using the first principle, find the derivative of $\tan \sqrt{x}$ w.r.t $x$

Answer 1

Hint: Here let’s use the formula of $\tan \left( A-B \right)$ before applying the limit and simplify. Then substitute the limit value to reach the final answer.

Complete step-by-step answer:
We know;
$\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
$\Rightarrow \tan A-\tan B=\tan (A-B)\left( 1+\tan A\tan B \right)$

If f(x) is any derivable function, then the derivative of f(x), by first principle, is given as:
$\dfrac{d\left( f(x) \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$

Applying the first principle for finding the derivative of $\tan \sqrt{x}$ :
$\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{tan\sqrt{x+h}-\tan \sqrt{x}}{h}$

Using the formula of $\tan (A-B)$ :
$\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{h}$

Adding and subtracting x in the denominator of the right hand side of the equation:$\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{x+h-x}$
\[\Rightarrow \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{{{\left( \sqrt{x+h} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}\]

Using formula: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
\[\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{\left( \sqrt{x+h}-\sqrt{x} \right)\left( \sqrt{x+h}+\sqrt{x} \right)}\]

And in our equation, if we put h=0, $\sqrt{x+h}-\sqrt{x}=0$.

We know;
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1$.

Using this to our equation, we get:
\[\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{\left( \sqrt{x+h}+\sqrt{x} \right)}\left( \dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)}{\left( \sqrt{x+h}-\sqrt{x} \right)} \right)\]
\[\dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{\left( \sqrt{x+h}+\sqrt{x} \right)}\times \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{tan\left( \sqrt{x+h}-\sqrt{x} \right)}{\left( \sqrt{x+h}-\sqrt{x} \right)} \right)\]
\[\therefore \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{\left( \sqrt{x+h}+\sqrt{x} \right)}\times 1\]
\[\Rightarrow \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+\tan \sqrt{x+h}\tan \sqrt{x} \right)}{\left( \sqrt{x+h}+\sqrt{x} \right)}\]

Finally putting the limit to the expression, we get:
\[\Rightarrow \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\dfrac{\left( 1+\tan \sqrt{x}\tan \sqrt{x} \right)}{\left( \sqrt{x}+\sqrt{x} \right)}\]
\[\Rightarrow \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\dfrac{\left( 1+{{\left( \tan \sqrt{x} \right)}^{2}} \right)}{2\sqrt{x}}\]

Using formula: $1+{{\tan }^{2}}A={{\sec }^{2}}A$ .
\[\therefore \dfrac{d\left( \tan \sqrt{x} \right)}{dx}=\dfrac{{{\sec }^{2}}\sqrt{x}}{2\sqrt{x}}\]

Hence, we can say that derivative of $\tan \sqrt{x}$ w.r.t $x$ is \[\dfrac{{{\sec }^{2}}\sqrt{x}}{2\sqrt{x}}\] .

Note: Other useful trigonometric formulas include:
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

And you are free to use any formula, try to figure out the method that will give you the answer in the least possible steps.

Point to remember: whenever you try to find the derivative avoid using the first principle until and unless it is mentioned in the question. Using the first principle might take longer as compared to the traditional method and may require a good knowledge related to different forms of formulas related to algebra, trigonometry, and also a bit of manipulation.

While solving limits, always keep a habit of checking the indeterminate form of the expression at the beginning of the question. There are a total of eight indeterminate forms that you should be knowing. Also, try to eliminate the particular term, which is causing the expression to be indeterminate.