What is the derivative of y=(x^2-6x+10)sqrt(25-x^2) ?

The derivative of y=(x^2-6x+10)sqrt(25-x^2) is (-3x^3+12x^2+40x-150)/(sqrt(25-x^2))

What is the first derivative of y=(x^2-6x+10)sqrt(25-x^2) ?

The first derivative of y=(x^2-6x+10)sqrt(25-x^2) is (-3x^3+12x^2+40x-150)/(sqrt(25-x^2))

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derivative of

y = (x^{2} - 6 x + 10) 25 - x^{2}

\frac{- 3 x ^{3} + 12 x ^{2} + 40 x - 150}{25 - x ^{2}}

Solution steps

\frac{d}{d x} ((x^{2} - 6 x + 10) 25 - x^{2})

Apply the Product Rule:

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

f = x^{2} - 6 x + 10, g = 25 - x^{2}

= \frac{d}{d x} (x^{2} - 6 x + 10) 25 - x^{2} + \frac{d}{d x} (25 - x^{2}) (x^{2} - 6 x + 10)

\frac{d}{d x} (x^{2} - 6 x + 10) = 2 x - 6

\frac{d}{d x} (25 - x^{2}) = - \frac{x}{25 - x ^{2}}

= (2 x - 6) 25 - x^{2} + (- \frac{x}{25 - x ^{2}}) (x^{2} - 6 x + 10)

Simplify

(2 x - 6) 25 - x^{2} + (- \frac{x}{25 - x ^{2}}) (x^{2} - 6 x + 10) : \frac{- 3 x ^{3} + 12 x ^{2} + 40 x - 150}{25 - x ^{2}}

= \frac{- 3 x ^{3} + 12 x ^{2} + 40 x - 150}{25 - x ^{2}}

\int 13 tan^{4} (x se) c^{6} x d x

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.1}} d x

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 1

\int \frac{x ^{2}}{( x ^{3} - 5 ) ^{3}} d x

x \to - 8 lim (\frac{x + 7}{x + 8})

What is the derivative of y=(x^2-6x+10)sqrt(25-x^2) ?
The derivative of y=(x^2-6x+10)sqrt(25-x^2) is (-3x^3+12x^2+40x-150)/(sqrt(25-x^2))
What is the first derivative of y=(x^2-6x+10)sqrt(25-x^2) ?
The first derivative of y=(x^2-6x+10)sqrt(25-x^2) is (-3x^3+12x^2+40x-150)/(sqrt(25-x^2))