What is the derivative of [x+(x+sin^2(x)^6]^3) ?

The derivative of [x+(x+sin^2(x)^6]^3) is 3(x+(x+sin^2(x))^6)^2(1+6(x+sin^2(x))^5(1+sin(2x)))

What is the first derivative of [x+(x+sin^2(x)^6]^3) ?

The first derivative of [x+(x+sin^2(x)^6]^3) is 3(x+(x+sin^2(x))^6)^2(1+6(x+sin^2(x))^5(1+sin(2x)))

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\frac{d}{d x} ([x + (x + sin^{2} (x))^{6}]^{3})

3 (x + (x + sin^{2} (x))^{6})^{2} (1 + 6 (x + sin^{2} (x))^{5} (1 + sin (2 x)))

Solution steps

\frac{d}{d x} ((x + (x + sin^{2} (x))^{6})^{3})

Apply the chain rule:

3 (x + (x + sin^{2} (x))^{6})^{2} \frac{d}{d x} (x + (x + sin^{2} (x))^{6})

= 3 (x + (x + sin^{2} (x))^{6})^{2} \frac{d}{d x} (x + (x + sin^{2} (x))^{6})

\frac{d}{d x} (x + (x + sin^{2} (x))^{6}) = 1 + 6 (x + sin^{2} (x))^{5} (1 + sin (2 x))

= 3 (x + (x + sin^{2} (x))^{6})^{2} (1 + 6 (x + sin^{2} (x))^{5} (1 + sin (2 x)))

\frac{\partial}{\partial y} (x e^{x y + 1})

\int_{- 1}^{x^{3}} 2 + t^{2} d t

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 9

(x - 1) ln (x), 1

\frac{\partial}{\partial x} (sin (x))

What is the derivative of [x+(x+sin^2(x)^6]^3) ?
The derivative of [x+(x+sin^2(x)^6]^3) is 3(x+(x+sin^2(x))^6)^2(1+6(x+sin^2(x))^5(1+sin(2x)))
What is the first derivative of [x+(x+sin^2(x)^6]^3) ?
The first derivative of [x+(x+sin^2(x)^6]^3) is 3(x+(x+sin^2(x))^6)^2(1+6(x+sin^2(x))^5(1+sin(2x)))