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In this lesson, we will learn how to find higher-order derivatives (d²y/dx²) of parametric equations by applying the chain rule.

Q1:

Given that π₯ = π‘ + 5 3 and π¦ = π‘ β 3 π‘ 2 , find d d 2 2 π¦ π₯ .

Q2:

Given that π₯ = 3 π‘ + 1 3 and π¦ = 3 π‘ β π‘ 2 , find d d 2 2 π¦ π₯ .

Q3:

Given that π₯ = 3 π‘ + 1 3 and π¦ = 5 π‘ β π‘ 2 , find d d 2 2 π¦ π₯ .

Q4:

Given that π₯ = 2 π 2 π‘ and π¦ = π‘ π β 2 π‘ , find d d 2 2 π¦ π₯ .

Q5:

Given that π₯ = π π‘ and π¦ = 4 π‘ π β π‘ , find d d 2 2 π¦ π₯ .

Q6:

Given that π₯ = π‘ + 1 2 and π¦ = π β 1 π‘ , find d d 2 2 π¦ π₯ .

Q7:

Given that π₯ = 2 π‘ + 4 2 and π¦ = 5 π β 4 5 π‘ , find d d 2 2 π¦ π₯ .

Q8:

Determine d d 2 2 π¦ π₯ , given that π₯ = 6 π l n 5 and π¦ = β 8 π 3 .

Q9:

Given that d d π§ π₯ = 5 π₯ β 6 and d d π¦ π₯ = 2 π₯ β 1 2 , determine d d 2 2 π§ π¦ at π₯ = 1 .

Q10:

Given that d d π§ π₯ = β 7 π₯ + 7 and d d π¦ π₯ = 3 π₯ β 1 2 , determine d d 2 2 π§ π¦ at π₯ = 0 .

Q11:

Given that d d π§ π₯ = β π₯ + 6 and d d π¦ π₯ = 2 π₯ + 5 2 , determine d d 2 2 π§ π¦ at π₯ = β 1 .

Q12:

If π₯ = 8 8 π§ s e c and β 5 π¦ = 7 8 π§ t a n , find d d 2 2 π¦ π₯ .

Q13:

If π₯ = 4 6 π§ s e c and β π¦ = 7 6 π§ t a n , find d d 2 2 π¦ π₯ .

Q14:

If π₯ = 2 5 π§ s e c and β 3 π¦ = 5 π§ t a n , find d d 2 2 π¦ π₯ .

Q15:

Find d d 2 2 π¦ π₯ if π₯ = β π 4 π and π¦ = β 2 π 4 .

Q16:

If π¦ = ( π₯ + 4 ) οΉ β 4 π₯ β 1 ο 2 and π§ = ( π₯ β 5 ) ( π₯ + 4 ) , find ( 2 π₯ β 1 ) π¦ π§ 3 2 2 d d .

Q17:

If π¦ = ( β π₯ + 2 ) οΉ 3 π₯ + 2 ο 2 and π§ = ( β π₯ + 2 ) ( π₯ + 3 ) , find ( β 2 π₯ β 1 ) π¦ π§ 3 2 2 d d .

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