What is the tangent of x(x^2-4x+5)^4,\at x=2 ?

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

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tangent of x(x^2-4x+5)^4,\at x=2

tangent of

x (x^{2} - 4 x + 5)^{4}, at x = 2

y = x

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, 2) : y = x

y = x

x \to 4 + lim (\frac{x + 2}{x - 4})

\int_{3}^{5} x \cdot \frac{x}{8} d x

\frac{8}{sin ( x ) + cos ( x )}

f (x) = 4 x^{2} - 5 x (- 1.9)

\int_{3}^{4} \frac{1}{( x - 3 ) ^{2}} d x

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