Question

Y= (secx+2)(tanx+3)
Derivatives

Answer 1

$\rm \implies y = (sec \: x + 2)(tan \: x + 3)$

Now differentiating both sides :-

$\rm \implies \dfrac{dy}{dx} =\dfrac{d \bigg((sec \: x + 2)(tan \: x + 3)\bigg)}{dx}$

Now we will use product rule :-

$\boxed{ \bf\dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx}}$

$\rm \implies \dfrac{dy}{dx} =(tan \: x + 3) \: \dfrac{d(sec \: x + 2)}{dx} + (sec \: x + 2) \: \dfrac{d (tan \: x + 3)}{dx}$

$\rm \implies \dfrac{dy}{dx} =(tan \: x + 3) \:(sec \: x \: tan \: x + 0) + (sec \: x + 2) \: {( {sec}^{2} \: x + 0)}$

$\rm \implies \dfrac{dy}{dx} =(tan \: x + 3) \:(sec \: x \: tan \: x) + (sec \: x + 2) \: {( {sec}^{2} \: x )}$

taking out sec x common :-

$\rm \implies \dfrac{dy}{dx} =sec \: x \bigg ((tan \: x + 3) \:( \: tan \: x) + (sec \: x + 2) \: {( {sec} \: x )} \bigg)$

$\rm \implies \dfrac{dy}{dx} =sec \: x \bigg ( {tan}^{2} \: x + 3 \: tan \: x + {sec}^{2} \: x + 2 {{sec} \: x } \bigg)$

Answer :

$\sf\dfrac{dy}{dx} =sec \: x \bigg ( {tan}^{2} \: x + 3 \: tan \: x + {sec}^{2} \: x + 2 {{sec} \: x } \bigg)$

____________________

Learn more :

d( c)/dx = 0 ....where c is constant

d( x^n)/dx= n x^( n-1)

d( e^x)/dx= e^x

d( ln(x))/dx= 1/x

d( sin(x))/dx = cosx

d( cos(x))/dx = -sinx

d( tan(x))/dx = sec²x

d( sec(x))/dx = sec x tan x

d( cosec (x))/dx = -cotx cosec x

d( cot (x))/dx = - cosec ² x

Answer 2

$\huge\underline{\bf \orange{AnSweR :}}$

Now differentiating both sides :-

=> dx/dy = dxd((secx+2)(tanx+3))

Now we will use product rule :-

=> dx/d(uv) =u dx/dv +v dx/du

=> (tanx+3) dx/d(secx+2) +(secx+2) dx/d(tanx+3)

=> dx/dy= (tanx+3)(secxtanx+0)+(secx+2)(sec 2 x+0)

=> dx/dy =(tanx+3)(secxtanx)+(secx+2)(sec 2 x)

taking out sec x common :-

=> dx/dy =secx((tanx+3)(tanx)+(secx+2)(secx))

=> dx/dy = Xsec(tan² x+3tanx+sec 2 x+2secx)

=> dx/dy =Xsec(tan²x+3tanx+sec 2 x+2secx)Ans.