What is the derivative of [x+(x+sin^2(x)^5]^4) ?

The derivative of [x+(x+sin^2(x)^5]^4) is 4(x+(x+sin^2(x))^5)^3(1+5(x+sin^2(x))^4(1+sin(2x)))

What is the first derivative of [x+(x+sin^2(x)^5]^4) ?

The first derivative of [x+(x+sin^2(x)^5]^4) is 4(x+(x+sin^2(x))^5)^3(1+5(x+sin^2(x))^4(1+sin(2x)))

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\frac{d}{d x} ([x + (x + sin^{2} (x))^{5}]^{4})

4 (x + (x + sin^{2} (x))^{5})^{3} (1 + 5 (x + sin^{2} (x))^{4} (1 + sin (2 x)))

Solution steps

\frac{d}{d x} ((x + (x + sin^{2} (x))^{5})^{4})

Apply the chain rule:

4 (x + (x + sin^{2} (x))^{5})^{3} \frac{d}{d x} (x + (x + sin^{2} (x))^{5})

= 4 (x + (x + sin^{2} (x))^{5})^{3} \frac{d}{d x} (x + (x + sin^{2} (x))^{5})

\frac{d}{d x} (x + (x + sin^{2} (x))^{5}) = 1 + 5 (x + sin^{2} (x))^{4} (1 + sin (2 x))

= 4 (x + (x + sin^{2} (x))^{5})^{3} (1 + 5 (x + sin^{2} (x))^{4} (1 + sin (2 x)))

y^{^{''}} - 4 y^{^{'}} + 20 y = 0

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

\frac{d}{d x} (2 x tanh (x))

- \frac{9}{( 6 t + 8 ) ^{\frac{3}{2}}}

x \to 1 - lim (\frac{4}{x ^{3} - 1})

What is the derivative of [x+(x+sin^2(x)^5]^4) ?
The derivative of [x+(x+sin^2(x)^5]^4) is 4(x+(x+sin^2(x))^5)^3(1+5(x+sin^2(x))^4(1+sin(2x)))
What is the first derivative of [x+(x+sin^2(x)^5]^4) ?
The first derivative of [x+(x+sin^2(x)^5]^4) is 4(x+(x+sin^2(x))^5)^3(1+5(x+sin^2(x))^4(1+sin(2x)))