What is the derivative of f(x)=8arcsin(x^4) ?

The derivative of f(x)=8arcsin(x^4) is (32x^3)/(sqrt(1-x^8))

What is the first derivative of f(x)=8arcsin(x^4) ?

The first derivative of f(x)=8arcsin(x^4) is (32x^3)/(sqrt(1-x^8))

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derivative of

f (x) = 8 arcsin (x^{4})

\frac{32 x ^{3}}{1 - x ^{8}}

Solution steps

\frac{d}{d x} (8 arcsin (x^{4}))

Take the constant out:

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

= 8 \frac{d}{d x} (arcsin (x^{4}))

Apply the chain rule:

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

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

= 8 \cdot \frac{1}{1 - ( x ^{4} ) ^{2}} \cdot 4 x^{3}

Simplify

8 \cdot \frac{1}{1 - ( x ^{4} ) ^{2}} \cdot 4 x^{3} : \frac{32 x ^{3}}{1 - x ^{8}}

= \frac{32 x ^{3}}{1 - x ^{8}}

x \to 3 + lim (∣ x - 3 ∣)

\int \frac{9}{1 - cos ^{2} ( x )} d x

\frac{d}{d x} (3 x^{2} - 18 x + 15)

\frac{4 s}{s ^{2} + 9}

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 7 y^{3}))

What is the derivative of f(x)=8arcsin(x^4) ?
The derivative of f(x)=8arcsin(x^4) is (32x^3)/(sqrt(1-x^8))
What is the first derivative of f(x)=8arcsin(x^4) ?
The first derivative of f(x)=8arcsin(x^4) is (32x^3)/(sqrt(1-x^8))