Exercise
Given
u ( x , y ) = e x y 2 u(x,y)=e^{xy^2} u ( x , y ) = e x y 2
f ( t ) = t cos t f(t)=t\cos t f ( t ) = t cos t
g ( t ) = t sin t g(t)=t\sin t g ( t ) = t sin t
U ( t ) = u ( f ( t ) , g ( t ) ) U(t)=u(f(t),g(t)) U ( t ) = u ( f ( t ) , g ( t ) )
Calculate the derivative
U ′ ( t ) U'(t) U ′ ( t )
Final Answer
Show final answer
U'(t)=t\sin te^{t^3\cos t\sin t}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t(\sin t+t\cos t)]
Solution
We will use the chain rule to calculate the derivative
U ′ ( t ) = u f ′ ⋅ f t ′ + u g ′ ⋅ g t ′ U'(t)=u'_f\cdot f'_t+u'_g\cdot g'_t U ′ ( t ) = u f ′ ⋅ f t ′ + u g ′ ⋅ g t ′
We have
U ( t ) = u ( f ( t ) , g ( t ) ) U(t)=u(f(t),g(t)) U ( t ) = u ( f ( t ) , g ( t ) )
And function u is
u ( x , y ) = e x y 2 u(x,y)=e^{xy^2} u ( x , y ) = e x y 2
We put it in the function above and get
u ( f , g ) = e f g 2 u(f,g)=e^{fg^2} u ( f , g ) = e f g 2
Hence, the partial derivatives of u are
u f ′ = g 2 e f g 2 u'_f=g^2e^{fg^2} u f ′ = g 2 e f g 2
u g ′ = 2 g f e f g 2 u'_g=2gfe^{fg^2} u g ′ = 2 g f e f g 2
Also, we calculate the derivatives of f and g
f t ′ = cos t − t sin t f'_t=\cos t -t\sin t f t ′ = cos t − t sin t
g t ′ = sin t + t cos t g'_t=\sin t+t\cos t g t ′ = sin t + t cos t
We put all the derivatives and get
U ′ ( t ) = u f ′ ⋅ f t ′ + u g ′ ⋅ g t ′ = U'(t)=u'_f\cdot f'_t+u'_g\cdot g'_t= U ′ ( t ) = u f ′ ⋅ f t ′ + u g ′ ⋅ g t ′ =
= g 2 e f g 2 ⋅ ( cos t − t sin t ) + 2 g f e f g 2 ⋅ ( sin t + t cos t ) = =g^2e^{fg^2}\cdot(\cos t -t\sin t)+2gfe^{fg^2}\cdot (\sin t+t\cos t)= = g 2 e f g 2 ⋅ ( cos t − t sin t ) + 2 g f e f g 2 ⋅ ( sin t + t cos t ) =
= g e f g 2 ⋅ [ g ( cos t − t sin t ) + 2 f ⋅ ( sin t + t cos t ) ] = =ge^{fg^2}\cdot[g(\cos t -t\sin t)+2f\cdot (\sin t+t\cos t)]= = g e f g 2 ⋅ [ g ( cos t − t sin t ) + 2 f ⋅ ( sin t + t cos t ) ] =
= t sin t e t cos t ( t sin t ) 2 ⋅ [ t sin t ( cos t − t sin t ) + 2 t cos t ⋅ ( sin t + t cos t ) ] = =t\sin te^{t\cos t{(t\sin t)}^2}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t\cdot (\sin t+t\cos t)]= = t sin t e t c o s t ( t s i n t ) 2 ⋅ [ t sin t ( cos t − t sin t ) + 2 t cos t ⋅ ( sin t + t cos t ) ] =
= t sin t e t 3 cos t sin t ⋅ [ t sin t ( cos t − t sin t ) + 2 t cos t ( sin t + t cos t ) ] =t\sin te^{t^3\cos t\sin t}\cdot[t\sin t(\cos t -t\sin t)+2t\cos t(\sin t+t\cos t)] = t sin t e t 3 c o s t s i n t ⋅ [ t sin t ( cos t − t sin t ) + 2 t cos t ( sin t + t cos t ) ]
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