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

The tangent of y=(x^2-15)^4,\at x=4 is y=32x-127

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

tangent of

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

y = 32 x - 127

Solution steps

Find the tangent point:

(4, 1)

Find the slope of

y = (x^{2} - 15)^{4} : \frac{d y}{d x} = 8 x (x^{2} - 15)^{3}

Compute the slope of

y = (x^{2} - 15)^{4}

x = 4 : m = 32

Find the line with slope m=

32

and passing through

(4, 1) : y = 32 x - 127

y = 32 x - 127

\frac{d}{d x} (ln (- cos (x^{2}) + 1))

\frac{d}{d x} (sin (\frac{x}{x + 1}))

x \to 7 lim ((x - 7)^{2} + x^{2})

x \to 0 + lim (ln (\frac{1}{x ^{4}}))

u (x) = x

What is the tangent of y=(x^2-15)^4,\at x=4 ?
The tangent of y=(x^2-15)^4,\at x=4 is y=32x-127