Question

consider the following.

v = sin(2s² - 3t²)

find the first partial derivatives.

vₛ = 4s cos(2s² - 3t²)

vₜ = -6t cos(2s² - 3t²)

find all the second partial derivatives.

Explanation:

Step1: Find \(v_{ss}\)

Differentiate \(v_s = 4s\cos(2s^{2}-3t^{2})\) with respect to \(s\).
Use the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = 4s\), \(v=\cos(2s^{2}-3t^{2})\).
\(u^\prime=4\), \(v^\prime=-4s\sin(2s^{2}-3t^{2})\)
\(v_{ss}=4\cos(2s^{2}-3t^{2})+4s\times(- 4s\sin(2s^{2}-3t^{2}))=4\cos(2s^{2}-3t^{2})-16s^{2}\sin(2s^{2}-3t^{2})\)

Step2: Find \(v_{st}\)

Differentiate \(v_s = 4s\cos(2s^{2}-3t^{2})\) with respect to \(t\).
Use the chain - rule. The derivative of \(\cos(2s^{2}-3t^{2})\) with respect to \(t\) is \(6t\sin(2s^{2}-3t^{2})\)
\(v_{st}=4s\times6t\sin(2s^{2}-3t^{2}) = 24st\sin(2s^{2}-3t^{2})\)

Step3: Find \(v_{ts}\)

Differentiate \(v_t=-6t\cos(2s^{2}-3t^{2})\) with respect to \(s\).
Use the chain - rule. The derivative of \(\cos(2s^{2}-3t^{2})\) with respect to \(s\) is \(-4s\sin(2s^{2}-3t^{2})\)
\(v_{ts}=-6t\times(-4s\sin(2s^{2}-3t^{2}))=24st\sin(2s^{2}-3t^{2})\)

Step4: Find \(v_{tt}\)

Differentiate \(v_t=-6t\cos(2s^{2}-3t^{2})\) with respect to \(t\).
Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=-6t\), \(v = \cos(2s^{2}-3t^{2})\)
\(u^\prime=-6\), \(v^\prime = 6t\sin(2s^{2}-3t^{2})\)
\(v_{tt}=-6\cos(2s^{2}-3t^{2})+(-6t)\times(6t\sin(2s^{2}-3t^{2}))=-6\cos(2s^{2}-3t^{2})-36t^{2}\sin(2s^{2}-3t^{2})\)

Answer:

\(v_{ss}=4\cos(2s^{2}-3t^{2})-16s^{2}\sin(2s^{2}-3t^{2})\)
\(v_{st}=24st\sin(2s^{2}-3t^{2})\)
\(v_{ts}=24st\sin(2s^{2}-3t^{2})\)
\(v_{tt}=-6\cos(2s^{2}-3t^{2})-36t^{2}\sin(2s^{2}-3t^{2})\)