Z=tan(y+ax) +(y-ax)^3/2∂^2z/∂x^2 =a^2 ∂^2z/∂y^2
Answers
Answer:
Given z = tan (y + ax) – √(y-ax)
Differentiate partially w.r.t.x
zx = sec2(y + ax) a – [1/2√(y-ax)]×-a
= a sec2 (y + ax) + a/2√(y-ax)
Differentiate partially w.r.t.x
zxx = a×2 sec2 (y + ax) tan (y + ax)×a + (a/2)(-½)(y-ax)-3/2×-a
= 2a2 sec2(y+ax) tan (y+ax) + a2/4(y-ax)3/2
Differentiate partially w.r.t.y
zy = sec2(y+ax) – 1/2√(y-ax)
zyy = 2 sec2(y+ax) tan (y+ax) + 1/4(y-ax)3/2
zxx– a2zyy = 2a2 sec2(y+ax) tan (y+ax) + a2/4(y-ax)3/2 – a2[2 sec2(y+ax) tan (y+ax) + 1/4(y-ax)3/2]
= 0
Answer:
Function of two or more variables with respect to one of those variables is equal to the partial derivative of the same function with respect to the other variable.
Step-by-step explanation:
From the above question,
They have given :
The partial derivative of a function of two or more variables with respect to one of those variables is equal to the partial derivative of the same function with respect to the other variable.
So,
solving for a:
a =
Therefore, the second partial derivative of z with respect to x is:
where a is given by
a =
Partial derivatives are a type of derivative which is used to measure the rate of change of a function with respect to one of its variables, while keeping the other variables constant.
In other words, the partial derivative of a function of two or more variables with respect to one of those variables is equal to the partial derivative of the same function with respect to the other variable.
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