What is the derivative of g(x)=(5-x)^2(2x-1)^5 ?

The derivative of g(x)=(5-x)^2(2x-1)^5 is -2(5-x)(2x-1)^5+10(2x-1)^4(5-x)^2

What is the first derivative of g(x)=(5-x)^2(2x-1)^5 ?

The first derivative of g(x)=(5-x)^2(2x-1)^5 is -2(5-x)(2x-1)^5+10(2x-1)^4(5-x)^2

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derivative of

g (x) = (5 - x)^{2} (2 x - 1)^{5}

- 2 (5 - x) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}

Solution steps

\frac{d}{d x} ((5 - x)^{2} (2 x - 1)^{5})

Apply the Product Rule:

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

f = (5 - x)^{2}, g = (2 x - 1)^{5}

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

\frac{d}{d x} ((5 - x)^{2}) = - 2 (5 - x)

\frac{d}{d x} ((2 x - 1)^{5}) = 10 (2 x - 1)^{4}

= (- 2 (5 - x)) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}

Simplify

= - 2 (5 - x) (2 x - 1)^{5} + 10 (2 x - 1)^{4} (5 - x)^{2}

\int_{- 8}^{10} \frac{x}{2} + 10 d x

t an g e n t g (x) = x^{2} - 8, at x = - 3

x^{2} y^{^{''}} + x y^{^{'}} + 4 y = 0

t an g e n t f (x) = x^{2} - 5 x + 5, (0, 5)

\frac{d}{d t} (e^{t} (cos (t)))

What is the derivative of g(x)=(5-x)^2(2x-1)^5 ?
The derivative of g(x)=(5-x)^2(2x-1)^5 is -2(5-x)(2x-1)^5+10(2x-1)^4(5-x)^2
What is the first derivative of g(x)=(5-x)^2(2x-1)^5 ?
The first derivative of g(x)=(5-x)^2(2x-1)^5 is -2(5-x)(2x-1)^5+10(2x-1)^4(5-x)^2