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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derivative of

[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)))

x \to 2 + lim (\frac{x ^{2} - 4}{x - 2})

\int tan^{5} (4 x) sec^{4} (4 x) d x

x \to - \infty lim (- x^{2})

\frac{1}{( s ^{2} - 2 s + 5 )}

f (x) = 25 - x, at x = 0

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)))