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Question
Math
Posted 2 months ago

Let f(x,y,z)=y2zf(x, y, z)=y^{2} z and g(t)=(t,t2,t3)g(t)=\left(t, t^{2}, t^{3}\right).
h(t)=f(g(t))h(2)=
\begin{array}{l}
h(t)=f(g(t)) \\
h^{\prime}(2)=
\end{array}
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To find h(t)h'(t), we need to apply the chain rule to the composite function h(t)=f(g(t))h(t) = f(g(t))
step 2
The chain rule states that if h(t)=f(g(t))h(t) = f(g(t)), then h(t)=f(g(t))g(t)h'(t) = f'(g(t)) \cdot g'(t)
step 3
We have f(x,y,z)=y2zf(x, y, z) = y^2z and g(t)=(t,t2,t3)g(t) = (t, t^2, t^3). Thus, g(t)=(1,2t,3t2)g'(t) = (1, 2t, 3t^2)
step 4
To find f(g(t))f'(g(t)), we differentiate f(x,y,z)f(x, y, z) with respect to xx, yy, and zz and evaluate at g(t)g(t)
step 5
We get fx=0\frac{\partial f}{\partial x} = 0, fy=2yz\frac{\partial f}{\partial y} = 2yz, and fz=y2\frac{\partial f}{\partial z} = y^2
step 6
Evaluating these derivatives at g(t)g(t) gives fy(g(t))=2t2t3=2t5\frac{\partial f}{\partial y}(g(t)) = 2t^2 \cdot t^3 = 2t^5 and fz(g(t))=(t2)2=t4\frac{\partial f}{\partial z}(g(t)) = (t^2)^2 = t^4
step 7
Now, f(g(t))=(0,2t5,t4)f'(g(t)) = \left(0, 2t^5, t^4\right)
step 8
Multiplying f(g(t))f'(g(t)) by g(t)g'(t) component-wise gives h(t)=(0,2t5,t4)(1,2t,3t2)h'(t) = \left(0, 2t^5, t^4\right) \cdot (1, 2t, 3t^2)
step 9
This results in h(t)=0+4t6+3t6=7t6h'(t) = 0 + 4t^6 + 3t^6 = 7t^6
step 10
Finally, evaluate h(t)h'(t) at t=2t = 2 to find h(2)=7(2)6h'(2) = 7(2)^6
step 11
Calculating the value gives h(2)=764=448h'(2) = 7 \cdot 64 = 448
[question number] Answer
h(2)=448h'(2) = 448
Key Concept
Chain Rule for Differentiation of Composite Functions
Explanation
To find the derivative of a composite function, we apply the chain rule, which involves taking the derivative of the outer function evaluated at the inner function and multiplying it by the derivative of the inner function.

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