Derivative of sin4x by First Principle [Limit Definition]

The derivative of sin4x is equal to 4cos4x.  In this post, we will find the derivative of sin4x by the first principle, that is, by the limit definition of derivatives.

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Derivative of sin4x

The limit definition (i.e., first principle) of derivatives tells us that the derivative of a function f(x) is given by the following limit:

$\dfrac{d}{dx}(f(x))$$=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$ …(I)

Derivative of Sin4x by First Principle

Question: What is the derivative of $\sin 4x$?

Answer: The derivative of sin4x is 4cos4x.

Explanation:

Step 1: We put $f(x)=\sin 4x$ in the above formula (I).

Step 2: Thus the derivative of sin4x by the first principle will be equal to

$\dfrac{d}{dx}(\sin 4x)$$=\lim\limits_{h\to 0} \dfrac{\sin4(x+h)-\sin 4x}{h}$

Step 3: Applying the formula $\sin a -\sin b$ $=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}$, we obtain that

$\dfrac{d}{dx}(\sin 4x)$$=\lim\limits_{h\to 0} \dfrac{1}{h} \cdot 2 \cos \dfrac{8x+4h}{2}\sin \frac{4h}{2}$

= $\lim\limits_{h \to 0} \frac{2}{h} \cdot \cos(4x+2h) \cdot \sin 2h$

= $4\lim\limits_{h \to 0} \cos(4x+2h)$ $\times \lim\limits_{h \to 0} \dfrac{\sin 2h}{2h}$

[Let $z=2h$. Then $z \to 0$ as $h \to 0$]

= $4 \cos(4x+0)$ $\times \lim\limits_{z \to 0} \dfrac{\sin z}{z}$

= $4 \cos4x \cdot 1$ as the limit of sinx/x is 1 when x tends to zero.

= $4\cos 4x$.

Conclusion: Thus, the derivative of sin4x is 4cos4x, obtained by the first principle of derivatives.

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Question-Answer on Derivative of Sin4x

Question: Find the derivative of sin4x at x=0.

Answer:

From the above, we know that the derivative of sin4x is 4cos4x. Thus, the derivative of sin4x at x=0 is equal to

$[\dfrac{d}{dx}(\sin 4x)]{x=0}$

$=[4\cos 4x]{x=0}$

$=4\cos 0$

$=4 \cdot 1$ as the value of cos0 is 1.

$=4$.

So the derivative of sin4x at x=0 is equal to 4.

FAQs

Q1: What is the derivative of sin4x?

Answer: The derivative of sin4x is 4cos4x.

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