What is the derivative of msin(2(mx+ny(x))) ?

The derivative of msin(2(mx+ny(x))) is mcos(2(mx+n{y}(x)))*2(m+n derivative of {y}(x))

What is the first derivative of msin(2(mx+ny(x))) ?

The first derivative of msin(2(mx+ny(x))) is mcos(2(mx+n{y}(x)))*2(m+n derivative of {y}(x))

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\frac{d}{d x} (m sin (2 (m x + n y (x))))

m cos (2 (m x + n y (x))) \cdot 2 (m + n \frac{d}{d x} (y (x)))

Solution steps

\frac{d}{d x} (m sin (2 (m x + n y (x))))

Take the constant out:

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

= m \frac{d}{d x} (sin (2 (m x + n y (x))))

Apply the chain rule:

cos (2 (m x + n y (x))) \frac{d}{d x} (2 (m x + n y (x)))

= cos (2 (m x + n y (x))) \frac{d}{d x} (2 (m x + n y (x)))

\frac{d}{d x} (2 (m x + n y (x))) = 2 (m + n \frac{d}{d x} (y (x)))

= m cos (2 (m x + n y (x))) \cdot 2 (m + n \frac{d}{d x} (y (x)))

x^{2} tan (x)

x \to + 0 lim (\frac{tan ( 6 x )}{sin ( 4 x )})

\int (x^{2} + 1) 5^{x^{3} + 3 x} d x

(4 + 3 x)^{^{'}}

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

What is the derivative of msin(2(mx+ny(x))) ?
The derivative of msin(2(mx+ny(x))) is mcos(2(mx+n{y}(x)))*2(m+n derivative of {y}(x))
What is the first derivative of msin(2(mx+ny(x))) ?
The first derivative of msin(2(mx+ny(x))) is mcos(2(mx+n{y}(x)))*2(m+n derivative of {y}(x))