What is the derivative of 5x^4(9-7x)^7 ?

The derivative of 5x^4(9-7x)^7 is 5(4x^3(9-7x)^7-49x^4(9-7x)^6)

What is the first derivative of 5x^4(9-7x)^7 ?

The first derivative of 5x^4(9-7x)^7 is 5(4x^3(9-7x)^7-49x^4(9-7x)^6)

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derivative of

5 x^{4} (9 - 7 x)^{7}

5 (4 x^{3} (9 - 7 x)^{7} - 49 x^{4} (9 - 7 x)^{6})

Solution steps

\frac{d}{d x} (5 x^{4} (9 - 7 x)^{7})

Take the constant out:

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

= 5 \frac{d}{d x} (x^{4} (9 - 7 x)^{7})

Apply the Product Rule:

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

f = x^{4}, g = (9 - 7 x)^{7}

= 5 (\frac{d}{d x} (x^{4}) (9 - 7 x)^{7} + \frac{d}{d x} ((9 - 7 x)^{7}) x^{4})

\frac{d}{d x} (x^{4}) = 4 x^{3}

\frac{d}{d x} ((9 - 7 x)^{7}) = - 49 (9 - 7 x)^{6}

= 5 (4 x^{3} (9 - 7 x)^{7} + (- 49 (9 - 7 x)^{6}) x^{4})

Simplify

= 5 (4 x^{3} (9 - 7 x)^{7} - 49 x^{4} (9 - 7 x)^{6})

\int_{- π}^{π} 1

f (x) = x^{3} - 3 x^{2} - 8 x + 9

x \to \infty lim (5 + \frac{2}{x})

\int_{2}^{3} (x - 2)^{3} d x

sec^{2} (x) d y + csc (y) d x = 0, y (\frac{π}{4}) = 0

What is the derivative of 5x^4(9-7x)^7 ?
The derivative of 5x^4(9-7x)^7 is 5(4x^3(9-7x)^7-49x^4(9-7x)^6)
What is the first derivative of 5x^4(9-7x)^7 ?
The first derivative of 5x^4(9-7x)^7 is 5(4x^3(9-7x)^7-49x^4(9-7x)^6)