What is the tangent of f(x)=(x^2-48)^4,\at x=7 ?

The tangent of f(x)=(x^2-48)^4,\at x=7 is y=56x-391

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tangent of f(x)=(x^2-48)^4,\at x=7

tangent of

f (x) = (x^{2} - 48)^{4}, at x = 7

y = 56 x - 391

Solution steps

Find the tangent point:

(7, 1)

Find the slope of

f (x) = (x^{2} - 48)^{4} : \frac{df ( x )}{d x} = 8 x (x^{2} - 48)^{3}

Compute the slope of

f (x) = (x^{2} - 48)^{4}

x = 7 : m = 56

Find the line with slope m=

56

and passing through

(7, 1) : y = 56 x - 391

y = 56 x - 391

\frac{d}{d x} (\frac{x}{3 x + 4 ln ( x )})

\frac{d}{d x} (x + 11)

\int_{0}^{\frac{3}{2}} x cos (π) x d x

\int 1 + v^{2} d v

2 (2 e^{x^{2}} x^{2} + e^{x^{2}})

What is the tangent of f(x)=(x^2-48)^4,\at x=7 ?
The tangent of f(x)=(x^2-48)^4,\at x=7 is y=56x-391