What is the tangent of f(x)=(2-4x^2)^5,\at x=1 ?

The tangent of f(x)=(2-4x^2)^5,\at x=1 is y=-640x+608

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tangent of f(x)=(2-4x^2)^5,\at x=1

tangent of

f (x) = (2 - 4 x^{2})^{5}, at x = 1

y = - 640 x + 608

Solution steps

Find the tangent point:

(1, - 32)

Find the slope of

f (x) = (2 - 4 x^{2})^{5} : \frac{df ( x )}{d x} = - 40 x (2 - 4 x^{2})^{4}

Compute the slope of

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

x = 1 : m = - 640

Find the line with slope m=

- 640

and passing through

(1, - 32) : y = - 640 x + 608

y = - 640 x + 608

y^{^{'}} + 8 (tan (8 x)) y = 3 cos (8 x)

f (x) = x^{3} - 4 x^{2} - 6 x + 8

\int \frac{1}{x} \cdot ln (x) d x

\int w d w

What is the tangent of f(x)=(2-4x^2)^5,\at x=1 ?
The tangent of f(x)=(2-4x^2)^5,\at x=1 is y=-640x+608