Question

Find the derivative of functions $cosx$, $cosecx$, $secx$, $cotx$ using first principle​​​​​​

• A. $-sinx,-cosecxcotx,secxtanx,-cose{c}^{2}x$
• B. $sinx,-cosecxcotx,secxtanx,-cose{c}^{2}x$
• C. $-sinx,-cosecxcotx,-secxtanx,cose{c}^{2}x$
• D. $-sinx,cosecxcotx,-secxtanx,cose{c}^{2}x$

Answer 1

The correct option is A $-sinx,-cosecxcotx,secxtanx,-cose{c}^{2}x$

(i) $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{cos(x+h)-cosx}{h}$
$={lim}_{h\to 0})\frac{cosxcosh-sinxsinh-cosx}{h}$
$={lim}_{h\to 0}\frac{cosx(cosh-1)-sinxsinh}{h}$
$={lim}_{h\to 0}\frac{cosx(cosh-1)}{h}-{lim}_{h\to 0}\frac{sinxsinh}{h}$
$=cosx{lim}_{h\to 0}\frac{cosh-1}{h}-sinx{lim}_{h\to 0}\frac{sinh}{h}$
$=cosx\times 0-sinx\times 1=-sinx$

(ii) $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{cosec(x+h)-cosecx}{h}$
$={lim}_{h\to 0}\frac{\frac{1}{sin(x+h)}-\frac{1}{sinx}}{h}$
$={lim}_{h\to 0}\frac{\frac{sinx-sin(x+h)}{sin(x+h)sinx}}{h}$
Using, $sinA-sinB=2\frac{cos(A+B)}{2}\frac{sin(A-B)}{2}$
$={lim}_{h\to 0}\frac{2cos\frac{(x+x+h)}{2}sin\frac{(x-x-h)}{2}}{hsin(x+h)sinx}$
$={lim}_{h\to 0}\frac{2cos\frac{(2x+h)}{2}sin\frac{(-h)}{2}}{hsin(x+h)sinx}$
$={lim}_{h\to 0}\frac{cos\frac{(2x+h)}{2}}{sin(x+h)sinx}{lim}_{h\to 0}\frac{-sin\left(h/2\right)}{\frac{h}{2}}$
Second limit has value unity, which is already derived in theory class.
$=\frac{cos\left(x\right)}{\left(sin\right(x\left)sinx\right)}(-1)=-cosecxcotx$

(iii) $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{sec(x+h)-secx}{h}$
$={lim}_{h\to 0}\frac{\frac{1}{cos(x+h)}-\frac{1}{cosx}}{h}$
$={lim}_{h\to 0}\left(\right(cosx-cos(x+h)\left)/\right(cos(x+h)cosx\left)\right)/h$
Using, $cosA-cosB=-2\frac{sin(A+B)}{2}\frac{sin(A-B)}{2}$
$={lim}_{h\to 0}-\frac{2sin\frac{(x+x+h)}{2}sin\frac{(x-x-h)}{2}}{hcos(x+h)cosx}$
$={lim}_{h\to 0}-\frac{2sin\frac{(2x+h)}{2}sin\frac{(-h)}{2}}{hcos(x+h)cosx}$
$={lim}_{h\to 0}-\frac{sin\frac{(2x+h)}{2}}{cos(x+h)cosx}{lim}_{h\to 0}\frac{-sin\left(h/2\right)}{\frac{h}{2}}$
Second limit has value unity, which is already derived in theory class.
$={lim}_{h\to 0}-\frac{sin\frac{(2x+h)}{2}}{cos(x+h)cosx}{lim}_{h\to 0}\frac{-sin\left(h/2\right)}{\frac{h}{2}}=\frac{sin\left(x\right)}{\left(cos\right(x\left)cosx\right)}(-1)=secxtanx$

(iv) $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$={lim}_{h\to 0}\frac{cot(x+h)-cotx}{h}$
$={lim}_{h\to 0}\frac{\frac{cos(x+h)}{sin(x+h)}-\frac{cosx}{sinx}}{h}$
$={lim}_{h\to 0}\frac{\frac{cos(x+h)sinx-sin(x+h)cosx}{sin(x+h)sinx}}{h}$
Using, $sin(A-B)=sinAcosB-cosAsinB$
$={lim}_{h\to 0}\frac{sin(x-x-h)}{hsin(x+h)sinx}$
$={lim}_{h\to 0}\frac{sin(-h)}{hsin(x+h)sinx}$
$={lim}_{h\to 0}\frac{1}{sin(x+h)sinx}{lim}_{h\to 0}\frac{-sin\left(h\right)}{h}$
Second limit has value unity, which is already derived in theory class.
$=\frac{1}{sinxsinx}(-1)=-cose{c}^{2}x$