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Sure! Let's go through the step-by-step process to differentiate the function [tex]\( w = \sin(xy) + y \ln(xz) + z \)[/tex] with respect to [tex]\( t \)[/tex], given that [tex]\( x = e^t \)[/tex], [tex]\( y = t^2 \)[/tex], and [tex]\( z = 1 \)[/tex].
1. Substitute the expressions for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex]:
Given:
[tex]\[ x = e^t, \quad y = t^2, \quad z = 1 \][/tex]
Substitute these into [tex]\( w \)[/tex]:
[tex]\[ w = \sin(xy) + y \ln(xz) + z = \sin(t^2 e^t) + t^2 \ln(e^t \cdot 1) + 1 \][/tex]
2. Simplify the expression:
[tex]\[ w = \sin(t^2 e^t) + t^2 \ln(e^t) + 1 \][/tex]
Since [tex]\(\ln(e^t) = t\)[/tex], the expression further simplifies:
[tex]\[ w = \sin(t^2 e^t) + t^2 t + 1 = \sin(t^2 e^t) + t^3 + 1 \][/tex]
3. Find the derivative [tex]\( \frac{d w}{d t} \)[/tex]:
Now, we need to differentiate [tex]\( w \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{d w}{d t} = \frac{d}{d t} \left( \sin(t^2 e^t) + t^3 + 1 \right) \][/tex]
4. Differentiate each term one by one:
- The derivative of [tex]\( \sin(t^2 e^t) \)[/tex]:
[tex]\[ \frac{d}{d t} \sin(t^2 e^t) \][/tex]
Apply the chain rule. Let [tex]\( u = t^2 e^t \)[/tex], so we need to use the chain rule [tex]\(\frac{d}{d t} \sin(u) = \cos(u) \cdot \frac{d u}{d t} \)[/tex]:
[tex]\[ \frac{d u}{d t} = \frac{d}{d t} (t^2 e^t) \][/tex]
Differentiate [tex]\( t^2 e^t \)[/tex] using the product rule:
[tex]\[ \frac{d}{d t} (t^2 e^t) = 2t e^t + t^2 e^t = e^t (2t + t^2) = t e^t (2 + t) \][/tex]
Thus,
[tex]\[ \frac{d}{d t} \sin(t^2 e^t) = \cos(t^2 e^t) \cdot t e^t (2 + t) \][/tex]
- The derivative of [tex]\( t^3 \)[/tex]:
[tex]\[ \frac{d}{d t} (t^3) = 3t^2 \][/tex]
- The derivative of the constant term [tex]\( 1 \)[/tex]:
[tex]\[ \frac{d}{d t} (1) = 0 \][/tex]
5. Combine all parts:
[tex]\[ \frac{d w}{d t} = \cos(t^2 e^t) \cdot t e^t (2 + t) + 3t^2 \][/tex]
Simplify:
[tex]\[ \frac{d w}{d t} = t e^t (2 + t) \cos(t^2 e^t) + 3t^2 \][/tex]
So the derivative [tex]\( \frac{d w}{d t} \)[/tex] is:
[tex]\[ \boxed{\frac{d w}{d t} = t^2 + 2t + (t^2 e^t + 2t e^t) \cos(t^2 e^t)} \][/tex]
1. Substitute the expressions for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex]:
Given:
[tex]\[ x = e^t, \quad y = t^2, \quad z = 1 \][/tex]
Substitute these into [tex]\( w \)[/tex]:
[tex]\[ w = \sin(xy) + y \ln(xz) + z = \sin(t^2 e^t) + t^2 \ln(e^t \cdot 1) + 1 \][/tex]
2. Simplify the expression:
[tex]\[ w = \sin(t^2 e^t) + t^2 \ln(e^t) + 1 \][/tex]
Since [tex]\(\ln(e^t) = t\)[/tex], the expression further simplifies:
[tex]\[ w = \sin(t^2 e^t) + t^2 t + 1 = \sin(t^2 e^t) + t^3 + 1 \][/tex]
3. Find the derivative [tex]\( \frac{d w}{d t} \)[/tex]:
Now, we need to differentiate [tex]\( w \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{d w}{d t} = \frac{d}{d t} \left( \sin(t^2 e^t) + t^3 + 1 \right) \][/tex]
4. Differentiate each term one by one:
- The derivative of [tex]\( \sin(t^2 e^t) \)[/tex]:
[tex]\[ \frac{d}{d t} \sin(t^2 e^t) \][/tex]
Apply the chain rule. Let [tex]\( u = t^2 e^t \)[/tex], so we need to use the chain rule [tex]\(\frac{d}{d t} \sin(u) = \cos(u) \cdot \frac{d u}{d t} \)[/tex]:
[tex]\[ \frac{d u}{d t} = \frac{d}{d t} (t^2 e^t) \][/tex]
Differentiate [tex]\( t^2 e^t \)[/tex] using the product rule:
[tex]\[ \frac{d}{d t} (t^2 e^t) = 2t e^t + t^2 e^t = e^t (2t + t^2) = t e^t (2 + t) \][/tex]
Thus,
[tex]\[ \frac{d}{d t} \sin(t^2 e^t) = \cos(t^2 e^t) \cdot t e^t (2 + t) \][/tex]
- The derivative of [tex]\( t^3 \)[/tex]:
[tex]\[ \frac{d}{d t} (t^3) = 3t^2 \][/tex]
- The derivative of the constant term [tex]\( 1 \)[/tex]:
[tex]\[ \frac{d}{d t} (1) = 0 \][/tex]
5. Combine all parts:
[tex]\[ \frac{d w}{d t} = \cos(t^2 e^t) \cdot t e^t (2 + t) + 3t^2 \][/tex]
Simplify:
[tex]\[ \frac{d w}{d t} = t e^t (2 + t) \cos(t^2 e^t) + 3t^2 \][/tex]
So the derivative [tex]\( \frac{d w}{d t} \)[/tex] is:
[tex]\[ \boxed{\frac{d w}{d t} = t^2 + 2t + (t^2 e^t + 2t e^t) \cos(t^2 e^t)} \][/tex]
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