Derivative of Exponential

Calculus Level 2

If y = 3 x 2 e ln y y = 3 - x^2 e^{\ln y} and f ( x ) = y f(x) = y , what is the value of f ( 2 ) ? f'(-2)?

13 25 \frac{13}{25} 14 25 \frac{14}{25} 12 25 \frac{12}{25} 11 25 \frac{11}{25}

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2 solutions

Daniel Ferreira
Feb 21, 2015

Por partes,

e ln y = α log e α = ln y ln α = ln y α = y e^{\ln y} = \alpha \\\\ \log_e \alpha = \ln y \\\\ \ln \alpha = \ln y \\\\ \boxed{\alpha = y }

Isto posto,

y = 3 x 2 e ln y y = 3 x 2 y y + x 2 y = 3 y ( 1 + x 2 ) = 3 y = 3 1 + x 2 f ( x ) = 3 1 + x 2 f ( x ) = ( 3 ) ( 1 + x 2 ) ( 3 ) ( 1 + x 2 ) ( 1 + x 2 ) 2 f ( x ) = 3 2 x ( 1 + x 2 ) 2 f ( 2 ) = 3 2 ( 2 ) ( 5 ) 2 f ( 2 ) = 12 25 y = 3 - x^2 \cdot e^{\ln y} \\\\ y = 3 - x^2 \cdot y \\\\y + x^2y = 3 \\\\ y(1 + x^2) = 3 \\\\ y = \frac{3}{1 + x^2} \\\\ f(x) = \frac{3}{1 + x^2} \\\\ f'(x) = \frac{(3)' \cdot (1 + x^2) - (3) \cdot (1 + x^2)'}{(1 + x^2)^2} \\\\ f'(x) = \frac{- 3 \cdot 2x}{(1 + x^2)^2} \\\\ f'(- 2) = \frac{- 3 \cdot 2 \cdot (- 2)}{(5)^2} \\\\ \boxed{f'(- 2) = \frac{12}{25}}

y = 3 x 2 e l n y y=3-x^{2}e^{lny}

y = 3 x 2 y y=3-x^{2}y

y + x 2 y = 3 y+x^{2}y=3

y ( 1 + x 2 ) = 3 y(1+x^{2})=3

y = 3 1 + x 2 y=\frac{3}{1+x^{2}}

f ( x ) = y = 3 1 + x 2 f(x)=y=\frac{3}{1+x^{2}}

f ( x ) = y = 6 x ( 1 + x 2 ) 2 f'(x)=y'=\frac{-6x}{(1+x^{2})^2}

f ( 2 ) = 6 ( 2 ) ( 1 + ( 2 ) 2 ) 2 f'(-2)=\frac{-6(-2)}{(1+(-2)^{2})^2}

f ( 2 ) = 12 ( 1 + 4 ) 2 f'(-2)=\frac{12}{(1+4)^2}

f ( 2 ) = 12 ( 5 ) 2 f'(-2)=\frac{12}{(5)^2}

f ( 2 ) = 12 25 f'(-2)=\boxed{\frac{12}{25}}

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