Fint dw/dt using the chain rule&nbsp;W=x²y³; x=t³ ;y=t²

To find dwdt using the chain rule, we first need to express w in terms of t. Since w=x2y3 and x=t3 and y=t2, we have:w=(t3)2(t2)3=t6·t6=t12Thus, we have w=t12 and we can now differentiate with respect to t using the chain rule:dwdt=ddt(t12)=ddt(t6·t6)Using the product rule, we can differentiate t6·t6 with respect to t:ddt(t6·t6)=ddt(t6)·t6+t6·ddt(t6)=6t5·t6+t6·6t5=12t11Therefore, we have:dwdt=ddt(t12)=12t11Thus, dwdt=12t11 using the chain rule.

Step by step solution to "Fint dw/dt using the chain ruleW=x²y³; x=t³ ;y=t²"

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Advanced Math

Answered question

2022-05-11

Fint dw/dt using the chain rule

W=x²y³; x=t³ ;y=t²

Answer & Explanation

Nick Camelot

Skilled2023-05-07Added 164 answers

To find

\frac{d w}{d t}

using the chain rule, we first need to express

w

in terms of

t

. Since

w = x^{2} y^{3}

and

x = t^{3}

and

y = t^{2}

, we have:

w = (t^{3})^{2} (t^{2})^{3} = t^{6} \cdot t^{6} = t^{12}

Thus, we have

w = t^{12}

and we can now differentiate with respect to

t

using the chain rule:

\frac{d w}{d t} = \frac{d}{d t} (t^{12}) = \frac{d}{d t} (t^{6} \cdot t^{6})

Using the product rule, we can differentiate

t^{6} \cdot t^{6}

with respect to

t

\frac{d}{d t} (t^{6} \cdot t^{6}) = \frac{d}{d t} (t^{6}) \cdot t^{6} + t^{6} \cdot \frac{d}{d t} (t^{6}) = 6 t^{5} \cdot t^{6} + t^{6} \cdot 6 t^{5} = 12 t^{11}

Therefore, we have:

\frac{d w}{d t} = \frac{d}{d t} (t^{12}) = 12 t^{11}

Thus,

\frac{d w}{d t} = 12 t^{11}

using the chain rule.

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