FIND THE DERIVATIVES USING FIRST PRINCIPLE (1)-sinx (2)-cosx (3)-tanx (4)- secx (5)-cosecx (6)-cotx

1) Here, we need to calculate the derivative of - sin x using the first principle. So,
Let $f\left(x\right)=-\sin x$. Then,
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-\sin \left(x+h\right)-\left(-\sin \left(x\right)\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-\sin \left(x+h\right)+\sin \left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(x\right)-\sin \left(x+h\right)}{h} \end{gathered}$$
Next, using the identity $SinA-SinB=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$. So,
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{x+x+h}{2}}\right)\sin \left({\displaystyle \frac{x-x-h}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{2x+h}{2}}\right)\sin \left({\displaystyle \frac{-h}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\cos \left({\displaystyle \frac{2x+h}{2}}\right).\underset{h\to 0}{\lim }\frac{2\sin \left({\displaystyle \frac{-h}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\cos \left({\displaystyle \frac{2x}{2}+\frac{h}{2}}\right).\underset{h\to 0}{\lim }\frac{-\left(\sin \left({\displaystyle \frac{-h}{2}}\right)\right)}{-\left({\displaystyle \frac{h}{2}}\right)} \\ =\left(-1\right)\underset{h\to 0}{\lim }\cos \left({\displaystyle \frac{2x}{2}+\frac{h}{2}}\right).\underset{h\to 0}{\lim }\frac{\left(\sin \left({\displaystyle \frac{-h}{2}}\right)\right)}{-\left({\displaystyle \frac{h}{2}}\right)} \end{gathered}$$
Further, using the property $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\left(-1\right)\cos \left({\displaystyle x+\frac{0}{2}}\right).1 \\ =-\cos x \end{gathered}$$

Therefore , $\frac{d\left(-\sin x\right)}{dx}=-\cos x$

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2) Here, we need to calculate the derivative of - tan x using the first principle. So,
Let $f\left(x\right)=-\tan x$. Then,
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-\tan \left(x+h\right)-\left(-\tan \left(x\right)\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-\tan \left(x+h\right)+\tan \left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\left(\frac{1}{h}\right)\left(\frac{-\sin \left(x+h\right)}{\cos \left(x+h\right)}+\frac{\sin x}{\cos x}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1}{h}\right)\left(\frac{-\sin \left(x+h\right)\cos x+\sin x\cos \left(x+h\right)}{h\cos \left(x+h\right)\cos x}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1}{h}\right)\left(\frac{\sin x\cos \left(x+h\right)-\sin \left(x+h\right)\cos x}{h\cos \left(x+h\right)\cos x}\right) \end{gathered}$$
Next, using the identity $Sin\left(x-y\right)=\sin x\cos y-\cos x\sin y$. So,
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\underset{h\to 0}{\lim }\left(\frac{\sin \left(x-\left(x+h\right)\right)}{h\cos \left(x+h\right)\cos x}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(x-x-h\right)}{h\cos \left(x+h\right)\cos x}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(-h\right)}{h\cos \left(x+h\right)\cos x}\right) \\ =\underset{h\to 0}{\lim }\frac{\sin \left(-h\right)}{h}.\underset{h\to 0}{\lim }\left(\frac{1}{\cos \left(x+h\right)\cos x}\right) \\ =\underset{h\to 0}{\lim }\frac{-\sin \left(-h\right)}{-h}.\underset{h\to 0}{\lim }\left(\frac{1}{\cos \left(x+h\right)\cos x}\right) \\ =\left(-1\right)\underset{h\to 0}{\lim }\frac{\sin \left(-h\right)}{-h}.\underset{h\to 0}{\lim }\left(\frac{1}{\cos \left(x+h\right)\cos x}\right) \end{gathered}$$
Further, using the property $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$
$$\begin{gathered} \frac{df\left(x\right)}{dx}=\left(-1\right).1.\left(\frac{1}{\cos \left(x+0\right)\cos x}\right) \\ =-1\left(\frac{1}{{\cos }^{2}x}\right) \\ =-{\sec }^{2}x \end{gathered}$$

Therefore , $\frac{d\left(-\tan x\right)}{dx}=-{\sec }^{2}x$

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