What is the tangent of f(x)=x^2sqrt(x^3+1),\at x=5 ?

The tangent of f(x)=x^2sqrt(x^3+1),\at x=5 is y=(1465)/(2sqrt(14))x+75sqrt(14)-(7325)/(2sqrt(14))

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tangent of f(x)=x^2sqrt(x^3+1),\at x=5

tangent of

f (x) = x^{2} x^{3} + 1, at x = 5

y = \frac{1465}{2 14} x + 7514 - \frac{7325}{2 14}

Solution steps

Find the tangent point:

(5, 7514)

Find the slope of

f (x) = x^{2} x^{3} + 1 : \frac{df ( x )}{d x} = \frac{7 x ^{4} + 4 x}{2 x ^{3} + 1}

Compute the slope of

f (x) = x^{2} x^{3} + 1

x = 5 : m = \frac{1465}{2 14}

Find the line with slope m=

\frac{1465}{2 14}

and passing through

(5, 7514) : y = \frac{1465}{2 14} x + 7514 - \frac{7325}{2 14}

y = \frac{1465}{2 14} x + 7514 - \frac{7325}{2 14}

\frac{4 s}{( s ^{2} + 16 )}

x \to 0 lim (18 - x^{2})

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

x \to \infty lim (\frac{8}{x !})

y = ln (x + 3)

What is the tangent of f(x)=x^2sqrt(x^3+1),\at x=5 ?
The tangent of f(x)=x^2sqrt(x^3+1),\at x=5 is y=(1465)/(2sqrt(14))x+75sqrt(14)-(7325)/(2sqrt(14))