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whats the answer you get for 4(a)(i),(ii) and (iii) ?? Just wanna compare..I dun have the ans :(
)
I don't have the 'd' symbol for partial derivatives, so I will use df/dx to mean the partial derivative with respect to x and df/dy as the partial derivative with respect to y.
I will also insert multiplicaation symbols in some cases to clarify the answers to the questions.
(i) If f(x,y) = x^4 + 6√y - 10, then df/dx = 4x^3 and df/dy = 3/√y.
(ii) For this one, there is no x or y, so df/dx = df/dy = 0.
If this one is suppose to be find df/ds and df/dt, the answers are ...
1st, let's rewrite f as f(s,t) = t^7 * ln(s^2) + 9 * t^-3 - s^(4/7) ...
df/ds = (t^7)(2s)/(s^2) + (4/7)(s^(-3/7)) = 2(t^7)/s + 4/(7*s^(3/7)) and
df/dt = (7t^6)ln(s^2) - 27(t^-4) = (7t^6)ln(s^2) - 27/(t^4).
(iii) Given that f(x,y) = cos(4/x)e^(x²y - 5y³), we have a product rule for x
and a simpler derivaitve for y.
It can be seen that the derivative of cos(4/x) with respect to x is -sin(4/x)(-4/x²),
which is 4sin(4/x)/x² and the derivative of e^(x²y - 5y³) with respect to x is
(2xy)e^(x²y - 5y³).
This makes df/dx = [cos(4/x)][(2xy)e^(x²y-5y³)] + [4*sin(4/x)/x²)][e^(x²y-5y³)].
This can be rewritten as 2xy*cos(4/x)*e^(x²y-5y³) + 4*sin(4/x)e^(x²y-5y³)/x².
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