What is the derivative of y(e^x-1) ?

The derivative of y(e^x-1) is (dy)/(dx)(e^x-1)+e^xy

What is the first derivative of y(e^x-1) ?

The first derivative of y(e^x-1) is (dy)/(dx)(e^x-1)+e^xy

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = e^{x} - 1

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

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

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

\int (\frac{x}{5})^{n} d x

\int_{- 1}^{1} 2^{x} d x

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

\frac{d}{d x} (8 x sin (x))

x \to - 3 + lim (\frac{x - 3}{x - 4})

What is the derivative of y(e^x-1) ?
The derivative of y(e^x-1) is (dy)/(dx)(e^x-1)+e^xy
What is the first derivative of y(e^x-1) ?
The first derivative of y(e^x-1) is (dy)/(dx)(e^x-1)+e^xy