JEE Main 2019 · 8 April · Evening Shift · Mathematics · Differentiation
D. 333333
⇒dydx=f′(f(f(x)))×f′(f(x))×f′(x)+2f(x)×f′(x)⇒ \dfrac{dy}{dx} = {f}^{'}\left(f\left(f\left(x\right)\right)\right) \times {f}^{'}\left(f\left(x\right)\right) \times {f}^{'}\left(x\right) + 2f\left(x\right) \times {f}^{'}\left(x\right)⇒dxdy=f′(f(f(x)))×f′(f(x))×f′(x)+2f(x)×f′(x)
⇒(dydx)x=1=f′(f(f(1)))×f′(f(1))×f′(1)+2f(1)×f′(1)⇒ {\left(\dfrac{dy}{dx}\right)}_{x = 1} = f'\left(f\left(f\left(1\right)\right)\right) \times {f}^{'}\left(f\left(1\right)\right) \times {f}^{'}\left(1\right) + 2f\left(1\right) \times {f}^{'}\left(1\right)⇒(dxdy)x=1=f′(f(f(1)))×f′(f(1))×f′(1)+2f(1)×f′(1)
=f′(f(1))×f′(1)×3+2×1×3= {f}^{'}\left(f\left(1\right)\right) \times {f}^{'}\left(1\right) \times 3 + 2 \times 1 \times 3=f′(f(1))×f′(1)×3+2×1×3
=f′(1)×3×3+6= {f}^{'}\left(1\right) \times 3 \times 3 + 6=f′(1)×3×3+6
=3×3×3+6= 3 \times 3 \times 3 + 6=3×3×3+6
=27+6= 27 + 6=27+6
=33= 33=33
