What is the derivative of 1-e^x+(e-1sin((pi/2)x)) ?

The derivative of 1-e^x+(e-1sin((pi/2)x)) is (pi(e-1))/2 cos((pix)/2)-e^x

What is the first derivative of 1-e^x+(e-1sin((pi/2)x)) ?

The first derivative of 1-e^x+(e-1sin((pi/2)x)) is (pi(e-1))/2 cos((pix)/2)-e^x

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

\frac{π ( e - 1 )}{2} cos (\frac{π x}{2}) - e^{x}

Solution steps

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

Simplify

1 - e^{x} + (e - 1) sin ((\frac{π}{2}) x) : 1 - e^{x} + (e - 1) sin (\frac{π x}{2})

= \frac{d}{d x} (1 - e^{x} + (e - 1) sin (\frac{π x}{2}))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) - \frac{d}{d x} (e^{x}) + \frac{d}{d x} ((e - 1) sin (\frac{π x}{2}))

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

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

\frac{d}{d x} ((e - 1) sin (\frac{π x}{2})) = \frac{π ( e - 1 )}{2} cos (\frac{π x}{2})

= 0 - e^{x} + \frac{π ( e - 1 )}{2} cos (\frac{π x}{2})

Simplify

= \frac{π ( e - 1 )}{2} cos (\frac{π x}{2}) - e^{x}

What is the derivative of 1-e^x+(e-1sin((pi/2)x)) ?
The derivative of 1-e^x+(e-1sin((pi/2)x)) is (pi(e-1))/2 cos((pix)/2)-e^x
What is the first derivative of 1-e^x+(e-1sin((pi/2)x)) ?
The first derivative of 1-e^x+(e-1sin((pi/2)x)) is (pi(e-1))/2 cos((pix)/2)-e^x