What is the tangent of f(x)=x^3sqrt(5-x^2),\at x=2 ?

The tangent of f(x)=x^3sqrt(5-x^2),\at x=2 is y=-4x+16

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tangent of f(x)=x^3sqrt(5-x^2),\at x=2

tangent of

f (x) = x^{3} 5 - x^{2}, at x = 2

y = - 4 x + 16

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

f (x) = x^{3} 5 - x^{2} : \frac{df ( x )}{d x} = \frac{- 4 x ^{4} + 15 x ^{2}}{5 - x ^{2}}

Compute the slope of

f (x) = x^{3} 5 - x^{2}

x = 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(2, 8) : y = - 4 x + 16

y = - 4 x + 16

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

\frac{\partial}{\partial x} (x^{3} + sin (3 y))

\int \frac{1}{x ^{2} + c} d x

\frac{2}{( s - 4 ) ^{3}}

\int_{- π}^{π} sin (k x) cos (n x) d x

What is the tangent of f(x)=x^3sqrt(5-x^2),\at x=2 ?
The tangent of f(x)=x^3sqrt(5-x^2),\at x=2 is y=-4x+16