What is the tangent of f(x)=(x^2-15)^7,\at x=4 ?

The tangent of f(x)=(x^2-15)^7,\at x=4 is y=56x-223

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tangent of f(x)=(x^2-15)^7,\at x=4

tangent of

f (x) = (x^{2} - 15)^{7}, at x = 4

y = 56 x - 223

Solution steps

Find the tangent point:

(4, 1)

Find the slope of

f (x) = (x^{2} - 15)^{7} : \frac{df ( x )}{d x} = 14 x (x^{2} - 15)^{6}

Compute the slope of

f (x) = (x^{2} - 15)^{7}

x = 4 : m = 56

Find the line with slope m=

56

and passing through

(4, 1) : y = 56 x - 223

y = 56 x - 223

x \frac{d y}{d x} + 2 y = 6 x^{2} y

\frac{d}{d x} (a cos^{3} (x))

y = x^{2} ln (3 x)

y^{^{'}} = \frac{( 1 - y ) ^{2}}{x + 1}, y (0) = 5

\frac{d}{d x} (ln (ln (sec (x))))

What is the tangent of f(x)=(x^2-15)^7,\at x=4 ?
The tangent of f(x)=(x^2-15)^7,\at x=4 is y=56x-223