Question

d/dx (1+2 tanx) (5+4cosx) differentiation
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Answer 1

Need to FinD :-

• To find the derivative of given function.

$\red{\frak{Given}}\bigg\{\sf \dfrac{d}{dx}(1+2tanx)(5+4cosx)$

Here we can use the product rule of differenciation . Say we have two functions u and v and we need to differentiate uv . Then its differenciation with respect to x , will be , Let y = uv ,

$\twoheadrightarrow\boxed{\sf \red{\dfrac{dy}{dx}= u\dfrac{dv}{dx}+v\dfrac{du}{dx} }}$

Here ,

• u = ( 1 + 2 tan x )
• v = ( 5 + 4 cosx ) .
• y = ( 1 + 2 tanx)(5 + 4 cos x)

$\sf:\implies \dfrac{dy}{dx}= \dfrac{d}{dx}( 1 + 2tanx)(5+4cosx) \\\\\sf:\implies \dfrac{dy}{dx}= (1+2tanx)\bigg[ \dfrac{d}{dx}(5+4cos x) \bigg] + (5+4cosx)\bigg[ \dfrac{d}{dx}(1+2tanx)\bigg] \\\\\sf:\implies \dfrac{dy}{dx}= (1+2tanx)\bigg[ \dfrac{d}{dx}(5) + \dfrac{d}{dx}(4cosx) \bigg] + (5+4cosx)\bigg[ \dfrac{d}{dx}(1) + \dfrac{d}{dx}(2tanx) \bigg]$

• Now we know that , derivative of ,

• cos x = -sin x
• tan x = sec² x
• Constant (C) = 0

• Also , if ,

$\sf\to\red{ \dfrac{dy}{dx}= \dfrac{d}{dx} \{c \: f(x) \} }$

Where c is a constant , then constant can come out in form of multiplication. That is ,

$\sf\to\red{ \dfrac{dy}{dx}= c\dfrac{d}{dx} f(x) }$

$\sf:\implies \dfrac{dy}{dx}= (1+2tanx)(0 + -4sinx)+(5+4cosx)(0+2sec^2x) \\\\\sf:\implies \dfrac{dy}{dx} = (1+2tanx)(-4sinx) + (5+4cosx)(2sec^2x) \\\\\sf:\implies \dfrac{dy}{dx} = -4sinx - 8tanx.sinx + 10 sec^2x + 8 cosx.sec^2x \\\\\sf:\implies \dfrac{dy}{dx} = -4sinx - \dfrac{8sin^2}{cosx} + \dfrac{10}{cos^2x} + \dfrac{8}{cosx} \\\\\sf:\implies \dfrac{dy}{dx}= -4sinx + 10sec^2x + \dfrac{8sin^2x}{cosx}+\dfrac{8}{cosx} \\\\\sf:\implies \dfrac{dy}{dx} = -4sinx + 10sec^2x +\dfrac{8(1-sin^2x)}{cosx} \\\\\sf:\implies \dfrac{dy}{dx}= -4sinx + 10 sec^2x + \dfrac{8cos^2x}{cosx} \\\\\sf:\implies \underset{\blue{\sf Required\ Derivative}}{\underbrace{\boxed{\pink{\frak{ \dfrac{dy}{dx} = -4sinx + 10sec^2x + 8cosx }}}}}$