Exercise
Given
z ( t , x , y ) = tan ( 3 t + 2 x 2 − y ) z(t,x,y)=\tan (3t+2x^2-y) z ( t , x , y ) = tan ( 3 t + 2 x 2 − y )
f ( t ) = 1 t f(t)=\frac{1}{t} f ( t ) = t 1
v ( t ) = t v(t)=\sqrt{t} v ( t ) = t
Z ( t ) = z ( t , f ( t ) , v ( t ) ) Z(t)=z(t,f(t),v(t)) Z ( t ) = z ( t , f ( t ) , v ( t ) )
Calculate the derivative
Z ′ ( t ) Z'(t) Z ′ ( t )
Final Answer
Show final answer
Z'(t)=\frac{3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}}}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}
Solution
We will use the chain rule to calculate the derivative
Z ′ ( t ) = z t ′ + z f ′ ⋅ f t ′ + z v ′ ⋅ v t ′ Z'(t)=z'_t+z'_f\cdot f'_t+z'_v\cdot v'_t Z ′ ( t ) = z t ′ + z f ′ ⋅ f t ′ + z v ′ ⋅ v t ′
We have
Z ( t ) = z ( t , f , v ) Z(t)=z(t,f,v) Z ( t ) = z ( t , f , v )
And function z is
z ( t , x , y ) = tan ( 3 t + 2 x 2 − y ) z(t,x,y)=\tan (3t+2x^2-y) z ( t , x , y ) = tan ( 3 t + 2 x 2 − y )
Hence, we get
z ( t , f , v ) = tan ( 3 t + 2 f 2 − v ) z(t,f,v)=\tan (3t+2f^2-v) z ( t , f , v ) = tan ( 3 t + 2 f 2 − v )
We calculate the partial derivatives of z.
z t ′ = 1 cos 2 ( 3 t + 2 f 2 − v ) ⋅ 3 = z'_t=\frac{1}{\cos^2(3t+2f^2-v)}\cdot 3= z t ′ = cos 2 ( 3 t + 2 f 2 − v ) 1 ⋅ 3 =
= 3 cos 2 ( 3 t + 2 f 2 − v ) =\frac{3}{\cos^2(3t+2f^2-v)} = cos 2 ( 3 t + 2 f 2 − v ) 3
z f ′ = 1 cos 2 ( 3 t + 2 f 2 − v ) ⋅ 4 f = z'_f=\frac{1}{\cos^2(3t+2f^2-v)}\cdot 4f= z f ′ = cos 2 ( 3 t + 2 f 2 − v ) 1 ⋅ 4 f =
= 4 f cos 2 ( 3 t + 2 f 2 − v ) =\frac{4f}{\cos^2(3t+2f^2-v)} = cos 2 ( 3 t + 2 f 2 − v ) 4 f
z v ′ = 1 cos 2 ( 3 t + 2 f 2 − v ) ⋅ ( − 1 ) = z'_v=\frac{1}{\cos^2(3t+2f^2-v)}\cdot (-1)= z v ′ = cos 2 ( 3 t + 2 f 2 − v ) 1 ⋅ ( − 1 ) =
= − 1 cos 2 ( 3 t + 2 f 2 − v ) =\frac{-1}{\cos^2(3t+2f^2-v)} = cos 2 ( 3 t + 2 f 2 − v ) − 1
And the partial derivatives of f and v.
f t ′ = − 1 t 2 f'_t=\frac{-1}{t^2} f t ′ = t 2 − 1
v t ′ = 1 2 t v'_t=\frac{1}{2\sqrt{t}} v t ′ = 2 t 1
We put the derivatives and get
Z ′ ( t ) = z t ′ + z f ′ ⋅ f t ′ + z v ′ ⋅ v t ′ = Z'(t)=z'_t+z'_f\cdot f'_t+z'_v\cdot v'_t= Z ′ ( t ) = z t ′ + z f ′ ⋅ f t ′ + z v ′ ⋅ v t ′ =
= 3 cos 2 ( 3 t + 2 f 2 − v ) + 4 f cos 2 ( 3 t + 2 f 2 − v ) ⋅ − 1 t 2 + − 1 cos 2 ( 3 t + 2 f 2 − v ) ⋅ 1 2 t = =\frac{3}{\cos^2(3t+2f^2-v)}+\frac{4f}{\cos^2(3t+2f^2-v)}\cdot \frac{-1}{t^2}+\frac{-1}{\cos^2(3t+2f^2-v)}\cdot \frac{1}{2\sqrt{t}}= = cos 2 ( 3 t + 2 f 2 − v ) 3 + cos 2 ( 3 t + 2 f 2 − v ) 4 f ⋅ t 2 − 1 + cos 2 ( 3 t + 2 f 2 − v ) − 1 ⋅ 2 t 1 =
= 3 cos 2 ( 3 t + 2 ( 1 t ) 2 − t ) + 4 1 t cos 2 ( 3 t + 2 ( 1 t ) 2 − t ) ⋅ − 1 t 2 + − 1 cos 2 ( 3 t + 2 ( 1 t ) 2 − t ) ⋅ 1 2 t = =\frac{3}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}+\frac{4\frac{1}{t}}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}\cdot \frac{-1}{t^2}+\frac{-1}{\cos^2(3t+2{(\frac{1}{t})}^2-\sqrt{t})}\cdot \frac{1}{2\sqrt{t}}= = cos 2 ( 3 t + 2 ( t 1 ) 2 − t ) 3 + cos 2 ( 3 t + 2 ( t 1 ) 2 − t ) 4 t 1 ⋅ t 2 − 1 + cos 2 ( 3 t + 2 ( t 1 ) 2 − t ) − 1 ⋅ 2 t 1 =
= 1 cos 2 ( 3 t + 2 t 2 − t ) ( 3 + 4 t ⋅ − 1 t 2 − 1 2 t ) = =\frac{1}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}(3+\frac{4}{t}\cdot \frac{-1}{t^2}-\frac{1}{2\sqrt{t}})= = cos 2 ( 3 t + t 2 2 − t ) 1 ( 3 + t 4 ⋅ t 2 − 1 − 2 t 1 ) =
= 1 cos 2 ( 3 t + 2 t 2 − t ) ( 3 + − 4 t 3 − 1 2 t ) = =\frac{1}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})}(3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}})= = cos 2 ( 3 t + t 2 2 − t ) 1 ( 3 + t 3 − 4 − 2 t 1 ) =
= 3 + − 4 t 3 − 1 2 t cos 2 ( 3 t + 2 t 2 − t ) =\frac{3+\frac{-4}{t^3}-\frac{1}{2\sqrt{t}}}{\cos^2(3t+\frac{2}{t^2}-\sqrt{t})} = cos 2 ( 3 t + t 2 2 − t ) 3 + t 3 − 4 − 2 t 1
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