What is the derivative of (x+(x+sin^2(x)^3)^4) ?

The derivative of (x+(x+sin^2(x)^3)^4) is 4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2(1+sin(2x)))

What is the first derivative of (x+(x+sin^2(x)^3)^4) ?

The first derivative of (x+(x+sin^2(x)^3)^4) is 4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2(1+sin(2x)))

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\frac{d}{d x} ((x + (x + sin^{2} (x))^{3})^{4})

4 (x + (x + sin^{2} (x))^{3})^{3} (1 + 3 (x + sin^{2} (x))^{2} (1 + sin (2 x)))

Solution steps

\frac{d}{d x} ((x + (x + sin^{2} (x))^{3})^{4})

Apply the chain rule:

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

= 4 (x + (x + sin^{2} (x))^{3})^{3} \frac{d}{d x} (x + (x + sin^{2} (x))^{3})

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

= 4 (x + (x + sin^{2} (x))^{3})^{3} (1 + 3 (x + sin^{2} (x))^{2} (1 + sin (2 x)))

\int 3 x + \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}} d x

y = x^{\frac{2}{3}} - x

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

\int t sinh (m t) d t

f (x) = \frac{3}{x + 1}, at x = - 2

What is the derivative of (x+(x+sin^2(x)^3)^4) ?
The derivative of (x+(x+sin^2(x)^3)^4) is 4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2(1+sin(2x)))
What is the first derivative of (x+(x+sin^2(x)^3)^4) ?
The first derivative of (x+(x+sin^2(x)^3)^4) is 4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2(1+sin(2x)))