What is the derivative of 0.5sin^2(x-x+1) ?

The derivative of 0.5sin^2(x-x+1) is sin(x)cos(x)-1

What is the first derivative of 0.5sin^2(x-x+1) ?

The first derivative of 0.5sin^2(x-x+1) is sin(x)cos(x)-1

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\frac{d}{d x} (0.5 sin^{2} (x) - x + 1)

sin (x) cos (x) - 1

Solution steps

\frac{d}{d x} (0.5 sin^{2} (x) - x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (0.5 sin^{2} (x)) - \frac{d x}{d x} + \frac{d}{d x} (1)

\frac{d}{d x} (0.5 sin^{2} (x)) = sin (x) cos (x)

\frac{d x}{d x} = 1

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

= sin (x) cos (x) - 1 + 0

Simplify

= sin (x) cos (x) - 1

\frac{4}{( s + 1 ) ^{2}}

t \cdot y^{^{'}} + t y = 1 - y

t^{2} \cdot e^{a \cdot t}

t^{2} y^{^{''}} + 19 t y^{^{'}} + 81 y = 0

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

What is the derivative of 0.5sin^2(x-x+1) ?
The derivative of 0.5sin^2(x-x+1) is sin(x)cos(x)-1
What is the first derivative of 0.5sin^2(x-x+1) ?
The first derivative of 0.5sin^2(x-x+1) is sin(x)cos(x)-1