What is (\partial)/(\partial x)((M)(1+y(M,x))) ?

The answer to (\partial)/(\partial x)((M)(1+y(M,x))) is M(0+y(M,1))

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(\partial)/(\partial x)((M)(1+y(M,x)))

\frac{\partial}{\partial x} ((M) (1 + y (M, x)))

M (0 + y (M, 1))

Solution steps

\frac{\partial}{\partial x} ((M) (1 + y (M, x)))

Treat

M, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= M \frac{\partial}{\partial x} (1 + y (M, x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= M (\frac{\partial}{\partial x} (1) + \frac{\partial}{\partial x} (y (M, x)))

\frac{\partial}{\partial x} (1) = 0

\frac{\partial}{\partial x} (y (M, x)) = y (M, 1)

= M (0 + y (M, 1))

\frac{x ^{2} + ( y - 1 ) ^{2}}{x ^{2}}

\frac{2500 s ( 20 s + 200 )}{( s + 1 ) ( s ^{2} + 100 s + 10000 )}

\frac{( x - 2 ) ( x - 4 )}{4}

- \frac{( - x + 2 ) ^{2}}{4}

- \frac{2 s ( - s ^{2} - 27 )}{( s ^{2} - 9 ) ^{3}}

What is (\partial)/(\partial x)((M)(1+y(M,x))) ?
The answer to (\partial)/(\partial x)((M)(1+y(M,x))) is M(0+y(M,1))