What is the derivative of (2x-7^4(x^2+x+1)^5) ?

The derivative of (2x-7^4(x^2+x+1)^5) is 8(2x-7)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-7)^4

What is the first derivative of (2x-7^4(x^2+x+1)^5) ?

The first derivative of (2x-7^4(x^2+x+1)^5) is 8(2x-7)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-7)^4

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\frac{d}{d x} ((2 x - 7)^{4} (x^{2} + x + 1)^{5})

8 (2 x - 7)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 7)^{4}

Solution steps

\frac{d}{d x} ((2 x - 7)^{4} (x^{2} + x + 1)^{5})

Apply the Product Rule:

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

f = (2 x - 7)^{4}, g = (x^{2} + x + 1)^{5}

= \frac{d}{d x} ((2 x - 7)^{4}) (x^{2} + x + 1)^{5} + \frac{d}{d x} ((x^{2} + x + 1)^{5}) (2 x - 7)^{4}

\frac{d}{d x} ((2 x - 7)^{4}) = 8 (2 x - 7)^{3}

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

= 8 (2 x - 7)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 7)^{4}

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

\frac{\partial}{\partial x} (sin (3 x) cos (6 y))

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

\int 2 x sec^{2} (x) d x

What is the derivative of (2x-7^4(x^2+x+1)^5) ?
The derivative of (2x-7^4(x^2+x+1)^5) is 8(2x-7)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-7)^4
What is the first derivative of (2x-7^4(x^2+x+1)^5) ?
The first derivative of (2x-7^4(x^2+x+1)^5) is 8(2x-7)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-7)^4