What is the derivative of f(x)=6sin^6(sqrt(x)) ?

The derivative of f(x)=6sin^6(sqrt(x)) is (18sin^5(sqrt(x))cos(sqrt(x)))/(sqrt(x))

What is the first derivative of f(x)=6sin^6(sqrt(x)) ?

The first derivative of f(x)=6sin^6(sqrt(x)) is (18sin^5(sqrt(x))cos(sqrt(x)))/(sqrt(x))

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derivative of

f (x) = 6 sin^{6} (x)

\frac{18 sin ^{5} ( x ) cos ( x )}{x}

Solution steps

\frac{d}{d x} (6 sin^{6} (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} (sin^{6} (x))

Apply the chain rule:

6 (sin (x))^{5} \frac{d}{d x} (sin (x))

= 6 (sin (x))^{5} \frac{d}{d x} (sin (x))

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

= 6 \cdot 6 (sin (x))^{5} \frac{cos ( x )}{2 x}

Simplify

6 \cdot 6 sin^{5} (x) \frac{cos ( x )}{2 x} : \frac{18 sin ^{5} ( x ) cos ( x )}{x}

= \frac{18 sin ^{5} ( x ) cos ( x )}{x}

\frac{\partial}{\partial t} (t^{8} e^{- x})

e^{x} (2 x + 2)

\int_{0}^{16} 8 x d x

k = 1 \sum \infty \frac{1}{1 + k ^{2}}

f (x) = x^{5} - 2 x^{3} + x - 1

What is the derivative of f(x)=6sin^6(sqrt(x)) ?
The derivative of f(x)=6sin^6(sqrt(x)) is (18sin^5(sqrt(x))cos(sqrt(x)))/(sqrt(x))
What is the first derivative of f(x)=6sin^6(sqrt(x)) ?
The first derivative of f(x)=6sin^6(sqrt(x)) is (18sin^5(sqrt(x))cos(sqrt(x)))/(sqrt(x))