Question

Find the derivatives of
I. 7 sec x+ 5 cos x
II. 5x^3-41

Answer 1

1st Question :-

Derivative of $\underline{ \sf{7 \sec(x) + 5 \cos(x) }}$

Solution:-

$\sf{\implies \dfrac{ d }{dx} [ 7 \sec(x) + 5 \cos(x) ] }$

$\sf{\implies \dfrac{ d }{dx} [ 7 \sec(x)] + \dfrac{d}{dx} [ 5 \cos(x) ] }$

$\underline{\sf{\implies a. u(x) + b. v(x) = a. u'(x) + b. v'(x)} }$

$\sf{\implies 7. \dfrac{ d }{dx} [ \sec(x)] + 5 . \dfrac{d}{dx} \cos(x) ] }$

$\underline{\sf{\implies \dfrac{ d }{dx} [ \sec(x) ] = \tan x . \sec x }}$

$\underline{\sf{\implies \dfrac{ d }{dx} [ \cos(x) ] = - \sin x }}$

$\sf{\implies 7 \sec(x). \tan (x) + 5 ( - \sin x) }$

• $\underline{\underline{\sf{\implies 7 \sec(x). \tan (x) - 5 \sin x }} }$

2nd Question :-

Derivative of $\underline{ \sf{ 5 {x}^{3} -41 }}$

Solution :-

$\sf{\implies \dfrac{ d }{dx} [ 5 {x}^{3} - 41 ] }$

$\underline{\sf{\implies \dfrac{ d }{dx} [41] = 0 }}$

$\underline{\sf{\implies \dfrac{ d }{dx} [{x}^{n}] = n. {x}^{n-1} }}$

$\sf{\implies [ 5 .3 {x}^{2} - 0 ] }$

• $\underline{\underline{\sf{\implies [ 15 {x}^{2} ] }}}$