derivative of 7x^4cot(x) — step-by-step solution

⚠Ⓘ 7x^4cot(x) is read as 7x^4(cos(x)/sin(x)): use cos(u)/sin(u) for cotangent, defined when sin(u) ≠ 0

1) Let's perform the differentiation:

\frac{d}{d x} (7 x^{4} \cdot \frac{\cos (x)}{\sin (x)})

For

u1

7 x^{4} \cdot \cos (x)

and

v2

\sin (x)

, use the product/quotient rule:

\frac{d}{d x} (\frac{u1}{v2}) = \frac{\frac{d}{d x} (u1) \cdot v2 - u1 \cdot \frac{d}{d x} (v2)}{{v2}^{2}}

2) Let's differentiate

u1

\frac{d}{d x} (7 x^{4} \cdot \cos (x))

7 \cdot \frac{d}{d x} (x^{4} \cdot \cos (x))

⇥For

U3

x^{4}

and

V4

\cos (x)

, use the product/quotient rule:

\frac{d}{d x} (U3 \cdot V4) = \frac{d}{d x} (U3) \cdot V4 + U3 \cdot \frac{d}{d x} (V4)

⇥2.1) Let's differentiate

U3

\frac{d}{d x} (x^{4})

⇥

4 x^{3}

⇥2.2) Let's differentiate

V4

\frac{d}{d x} (\cos (x))

⇥

- \sin (x)

⇥2.3) You can evaluate the resulted derivative (

U3 \cdot V4

)':

4 x^{3} \cdot \cos (x) + x^{4} \cdot (- \sin (x))

⇥

4 x^{3} \cdot \cos (x) - x^{4} \cdot \sin (x)

⇥2.4) You can then resume the derivation:

7 \cdot (4 x^{3} \cdot \cos (x) - x^{4} \cdot \sin (x))

28 x^{3} \cdot \cos (x) + 7 \cdot (- x^{4}) \cdot \sin (x)

28 x^{3} \cdot \cos (x) - 7 x^{4} \cdot \sin (x)

3) Let's differentiate

v2

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

\cos (x)

4) You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

\frac{(28 x^{3} \cdot \cos (x) - 7 x^{4} \cdot \sin (x)) \cdot \sin (x) - 7 x^{4} \cdot \cos (x) \cdot \cos (x)}{{\sin (x)}^{2}}

\frac{(28 x^{3} \cdot \cos (x) - 7 x^{4} \cdot \sin (x)) \cdot \sin (x) - 7 x^{4} \cdot {\cos (x)}^{2}}{{\sin (x)}^{2}}

\frac{(28 x^{3} \cdot \cos (x) \cdot \sin (x) - 7 x^{4} \cdot {\sin (x)}^{2}) - 7 x^{4} \cdot {\cos (x)}^{2}}{{\sin (x)}^{2}}

\frac{7 x^{3} \cdot (4 \cdot \cos (x) \cdot \sin (x) - x \cdot {\sin (x)}^{2} - x \cdot {\cos (x)}^{2})}{{\sin (x)}^{2}}

▶Derivative:

\frac{7 x^{3} \cdot (4 \cdot \cos (x) \cdot \sin (x) - x)}{{\sin (x)}^{2}}