Question

find the derivative of √tan x by first principle ( no spam please)​

Answer 1

Step-by-step explanation:

We will use this formula for $ \tan \sqrt {x + h} - \tan \sqrt x $ . We will find the limits separately first and put them into the main equation. Thus, the derivative of $ \tan \sqrt x $ w.r.t. $ x $ using the first principle is $ \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x $ .

Answer 2

Answer:

I hope it is helpful

Step-by-step explanation:

We have f(x)=tanx−−√ , therefore f(x+h)=tanx+h−−−−−√

Now, we know that by the first principle, the derivative can is given by the following formula:

f′(x)=limh→0f(x+h)−f(x)h

Now, we will put f(x)=tanx−−√ and f(x+h)=tanx+h−−−−−√ in this formula.

⇒f′(x)=limh→0tanx+h−−−−−√−tanx−−√h

We know that tanA−tanB=tan(A−B)(1+tanAtanB) . We will use this formula for tanx+h−−−−−√−tanx−−√ .

We will put tanx+h−−−−−√−tanx−−√=tan(x+h−−−−−√−x−−√)(1+tanx+h−−−−−√tanx−−√)

⇒f′(x)=limh→0tan(x+h−−−−−√−x−−√)(1+tanx+h−−−−−√tanx−−√)h

Now we will separate the terms

⇒f′(x)=limh→0tan(x+h−−−−−√−x−−√)hlimh→0(1+tanx+h−−−−−√tanx−−√)

We will find the limits separately first and put them into the main equation.

First, we will solve limh→0tan(x+h−−−−−√−x−−√)h

We can write h=(x+h−−−−−√−x−−√)(x+h−−−−−√+x−−√)

limh→0tan(x+h−−−−−√−x−−√)h=limh→0tan(x+h−−−−−√−x−−√)(x+h−−−−−√−x−−√)(x+h−−−−−√+x−−√)=limh→0tan(x+h−−−−−√−x−−√)(x+h−−−−−√−x−−√)limh→01(x+h−−−−−√+x−−√)=1×12x−−√=12x−−√

Now, we will solve the second term limh→0(1+tanx+h−−−−−√tanx−−√)

limh→0(1+tanx+h−−−−−√tanx−−√)=(1+tanx−−√tanx−−√)=1+tan2x−−√=sec2x−−√

Now, we will put both these values in the main equation

⇒f′(x)=limh→0tan(x+h−−−−−√−x−−√)hlimh→0(1+tanx+h−−−−−√tanx−−√)⇒f′(x)=12x−−√sec2x−−√

Thus, the derivative of tanx−−√ w.r.t. x using the first principle is 12x−−√sec2x−−√ .

So, the correct answer is “12x−−√sec2x−−√ .”.

Note: There is a second method to find the derivative of tanx−−√ w.r.t. x . This method is called the chain rule. In this method we will find the derivative of the main function which is whose derivative is sec2y . Here, we have taken x−−√=y . Now, we will find the derivative of x−−√ which is 12x−−√ . Now, we will multiply both these derivatives and get the final answer as 12x−−√sec2x−−√ . This can also be written as:

ddxtanx−−√=12x−−√sec2x−−√