Exercise
Given
v ( x , y ) = x y v(x,y)=\frac{x}{y} v ( x , y ) = y x
f ( t ) = t e 2 t f(t)=te^{2t} f ( t ) = t e 2 t
g ( t ) = ln ( t 2 + ln ( 5 t ) ) g(t)=\ln (t^2+\ln (5t)) g ( t ) = ln ( t 2 + ln ( 5 t ) )
V ( t ) = v ( f ( t ) , g ( t ) ) V(t)=v(f(t),g(t)) V ( t ) = v ( f ( t ) , g ( t ) )
Calculate the derivative
V ′ ( t ) V'(t) V ′ ( t )
Final Answer
Show final answer
V'(t)=\frac{1}{\ln (t^2+\ln (5t))}[(e^{2t}+t\cdot 2e^{2t})-\frac{te^{2t}}{t^2+\ln (5t)}(2t+\frac{1}{t})]
Solution
We will use the chain rule to calculate the derivative
V ′ ( t ) = v f ′ ⋅ f t ′ + v g ′ ⋅ g t ′ V'(t)=v'_f\cdot f'_t+v'_g\cdot g'_t V ′ ( t ) = v f ′ ⋅ f t ′ + v g ′ ⋅ g t ′
We have
V ( t ) = v ( f ( t ) , g ( t ) ) V(t)=v(f(t),g(t)) V ( t ) = v ( f ( t ) , g ( t ) )
And function v is
v ( x , y ) = x y v(x,y)=\frac{x}{y} v ( x , y ) = y x
We put it in function u and get
u ( f , g ) = f g u(f,g)=\frac{f}{g} u ( f , g ) = g f
We calculate the patial derivatives of v.
v f ′ = 1 g v'_f=\frac{1}{g} v f ′ = g 1
v g ′ = − f g 2 v'_g=\frac{-f}{g^2} v g ′ = g 2 − f
Also, we calculate the derivatives of f and g.
f t ′ = e 2 t + t ⋅ 2 e 2 t f'_t=e^{2t}+t\cdot 2e^{2t} f t ′ = e 2 t + t ⋅ 2 e 2 t
g t ′ = 1 t 2 + ln ( 5 t ) ( 2 t + 1 5 t ⋅ 5 ) = g'_t=\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{5t}\cdot 5)= g t ′ = t 2 + ln ( 5 t ) 1 ( 2 t + 5 t 1 ⋅ 5 ) =
= 1 t 2 + ln ( 5 t ) ( 2 t + 1 t ) =\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t}) = t 2 + ln ( 5 t ) 1 ( 2 t + t 1 )
We put the derivatives in V’ and get
V ′ ( t ) = v f ′ ⋅ f t ′ + v g ′ ⋅ g t ′ = V'(t)=v'_f\cdot f'_t+v'_g\cdot g'_t= V ′ ( t ) = v f ′ ⋅ f t ′ + v g ′ ⋅ g t ′ =
= 1 g ⋅ ( e 2 t + t ⋅ 2 e 2 t ) + − f g 2 ⋅ 1 t 2 + ln ( 5 t ) ( 2 t + 1 t ) = =\frac{1}{g}\cdot(e^{2t}+t\cdot 2e^{2t})+\frac{-f}{g^2}\cdot\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t})= = g 1 ⋅ ( e 2 t + t ⋅ 2 e 2 t ) + g 2 − f ⋅ t 2 + ln ( 5 t ) 1 ( 2 t + t 1 ) =
= 1 ln ( t 2 + ln ( 5 t ) ) ⋅ ( e 2 t + t ⋅ 2 e 2 t ) + − t e 2 t ln 2 ( t 2 + ln ( 5 t ) ) ⋅ 1 t 2 + ln ( 5 t ) ( 2 t + 1 t ) = =\frac{1}{\ln (t^2+\ln (5t))}\cdot(e^{2t}+t\cdot 2e^{2t})+\frac{-te^{2t}}{\ln^2 (t^2+\ln (5t))}\cdot\frac{1}{t^2+\ln (5t)}(2t+\frac{1}{t})= = ln ( t 2 + ln ( 5 t ) ) 1 ⋅ ( e 2 t + t ⋅ 2 e 2 t ) + ln 2 ( t 2 + ln ( 5 t ) ) − t e 2 t ⋅ t 2 + ln ( 5 t ) 1 ( 2 t + t 1 ) =
= 1 ln ( t 2 + ln ( 5 t ) ) [ ( e 2 t + t ⋅ 2 e 2 t ) − t e 2 t t 2 + ln ( 5 t ) ( 2 t + 1 t ) ] =\frac{1}{\ln (t^2+\ln (5t))}[(e^{2t}+t\cdot 2e^{2t})-\frac{te^{2t}}{t^2+\ln (5t)}(2t+\frac{1}{t})] = ln ( t 2 + ln ( 5 t ) ) 1 [ ( e 2 t + t ⋅ 2 e 2 t ) − t 2 + ln ( 5 t ) t e 2 t ( 2 t + t 1 ) ]
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