Find all the second order partial derivatives of g (x, y) = x Superscript 4 Baseline y + 3 sine (y) + 5 y cosine (x)

Step-by-step explanation:

Given the function

before we can get its second order partial derivative, we need to get its first order first. The first order are δg/δx and δg/δy

δg/δx  

δg/δy =

The second derivatives are δ²g/δy²,  δ²g/δx², δ²g/ δyδx or δ²g/ δxδy

δ²g/δy² =  δ/δy (δg/δy) = δ/δy()  

δ²g/δy² = -3siny

Similarly δ²g/δx² =  δ/δx (δg/δx) = δ/δx()

δ²g/δx² = 12x²y-5ycosx

δ²g/ δyδx =   δ/δy (δg/δx) =   δ/δy ()

δ²g/ δyδx  = 4x³ - 5sinx = δ²g/ δxδy ( for continuous function)

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°