What is the derivative of 5y^3+(x-2^2-4-e^{x-2y}) ?

The derivative of 5y^3+(x-2^2-4-e^{x-2y}) is 2x-4+15y^2(dy)/(dx)-e^{x-2y}(1-2(dy)/(dx))

What is the first derivative of 5y^3+(x-2^2-4-e^{x-2y}) ?

The first derivative of 5y^3+(x-2^2-4-e^{x-2y}) is 2x-4+15y^2(dy)/(dx)-e^{x-2y}(1-2(dy)/(dx))

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\frac{d}{d x} (5 y^{3} + (x - 2)^{2} - 4 - e^{x - 2 y})

2 x - 4 + 15 y^{2} \frac{d y}{d x} - e^{x - 2 y} (1 - 2 \frac{d y}{d x})

Solution steps

\frac{d}{d x} (5 y^{3} + (x - 2)^{2} - 4 - e^{x - 2 y})

Simplify

5 y^{3} + (x - 2)^{2} - 4 - e^{x - 2 y} : x^{2} - 4 x + 5 y^{3} - e^{x - 2 y}

= \frac{d}{d x} (x^{2} - 4 x + 5 y^{3} - e^{x - 2 y})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) - \frac{d}{d x} (4 x) + \frac{d}{d x} (5 y^{3}) - \frac{d}{d x} (e^{x - 2 y})

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (5 y^{3}) = 15 y^{2} \frac{d y}{d x}

\frac{d}{d x} (e^{x - 2 y}) = e^{x - 2 y} (1 - 2 \frac{d y}{d x})

= 2 x - 4 + 15 y^{2} \frac{d y}{d x} - e^{x - 2 y} (1 - 2 \frac{d y}{d x})

y^{^{''}} + 4 y = - 12 cos (2 x) - 4 sin (2 x)

\frac{\partial}{\partial y} (x y ln (x^{2} + y^{2}))

\int \frac{2}{4 - x ^{2}} d x

\int \frac{( x + 2 )}{( x + 1 )} d x

y^{^{'}} (t) = (y + 1) \cdot (y - 3), y (0) = 1

What is the derivative of 5y^3+(x-2^2-4-e^{x-2y}) ?
The derivative of 5y^3+(x-2^2-4-e^{x-2y}) is 2x-4+15y^2(dy)/(dx)-e^{x-2y}(1-2(dy)/(dx))
What is the first derivative of 5y^3+(x-2^2-4-e^{x-2y}) ?
The first derivative of 5y^3+(x-2^2-4-e^{x-2y}) is 2x-4+15y^2(dy)/(dx)-e^{x-2y}(1-2(dy)/(dx))