What is the derivative of x+sin^2(x/3-8) ?

The derivative of x+sin^2(x/3-8) is 2/3 sin(x/3)cos(x/3)+1

What is the first derivative of x+sin^2(x/3-8) ?

The first derivative of x+sin^2(x/3-8) is 2/3 sin(x/3)cos(x/3)+1

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\frac{d}{d x} (x + sin^{2} (\frac{x}{3}) - 8)

\frac{2}{3} sin (\frac{x}{3}) cos (\frac{x}{3}) + 1

Solution steps

\frac{d}{d x} (x + sin^{2} (\frac{x}{3}) - 8)

Apply the Sum/Difference Rule:

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

= \frac{d x}{d x} + \frac{d}{d x} (sin^{2} (\frac{x}{3})) - \frac{d}{d x} (8)

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

\frac{d}{d x} (sin^{2} (\frac{x}{3})) = \frac{2}{3} sin (\frac{x}{3}) cos (\frac{x}{3})

\frac{d}{d x} (8) = 0

= 1 + \frac{2}{3} sin (\frac{x}{3}) cos (\frac{x}{3}) - 0

Simplify

= \frac{2}{3} sin (\frac{x}{3}) cos (\frac{x}{3}) + 1

\frac{d}{d x} (- 3^{x})

\int_{0}^{π} sin^{6} (x) d x

2 y^{^{''}} - 2 y^{^{'}} + y = 0

\int \frac{1}{t} d t

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

What is the derivative of x+sin^2(x/3-8) ?
The derivative of x+sin^2(x/3-8) is 2/3 sin(x/3)cos(x/3)+1
What is the first derivative of x+sin^2(x/3-8) ?
The first derivative of x+sin^2(x/3-8) is 2/3 sin(x/3)cos(x/3)+1