What is the derivative of 8x^2sin(xtan(x)) ?

The derivative of 8x^2sin(xtan(x)) is 8(2xsin(x)tan(x)+sin(x)(1+sec^2(x))x^2)

What is the first derivative of 8x^2sin(xtan(x)) ?

The first derivative of 8x^2sin(xtan(x)) is 8(2xsin(x)tan(x)+sin(x)(1+sec^2(x))x^2)

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\frac{d}{d x} (8 x^{2} sin (x) tan (x))

8 (2 x sin (x) tan (x) + sin (x) (1 + sec^{2} (x)) x^{2})

Solution steps

\frac{d}{d x} (8 x^{2} sin (x) tan (x))

Take the constant out:

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

= 8 \frac{d}{d x} (x^{2} sin (x) tan (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = sin (x) tan (x)

= 8 (\frac{d}{d x} (x^{2}) sin (x) tan (x) + \frac{d}{d x} (sin (x) tan (x)) x^{2})

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

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

= 8 (2 x sin (x) tan (x) + sin (x) (1 + sec^{2} (x)) x^{2})

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

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0, y (1) = 0.4, y^{^{'}} (1) = 0

y d x - (x + y) d y = 0

(5, - 5), (- 3, 5)

y^{^{'''}} - 7 y^{^{''}} + 8 y^{^{'}} + 16 y = 0

What is the derivative of 8x^2sin(xtan(x)) ?
The derivative of 8x^2sin(xtan(x)) is 8(2xsin(x)tan(x)+sin(x)(1+sec^2(x))x^2)
What is the first derivative of 8x^2sin(xtan(x)) ?
The first derivative of 8x^2sin(xtan(x)) is 8(2xsin(x)tan(x)+sin(x)(1+sec^2(x))x^2)