What is the tangent of f(x)=5x^2,\at x=10 ?

The tangent of f(x)=5x^2,\at x=10 is y=100x-500

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tangent of f(x)=5x^2,\at x=10

tangent of

f (x) = 5 x^{2}, at x = 10

y = 100 x - 500

Solution steps

Find the tangent point:

(10, 500)

Find the slope of

f (x) = 5 x^{2} : \frac{df ( x )}{d x} = 10 x

Compute the slope of

f (x) = 5 x^{2}

x = 10 : m = 100

Find the line with slope m=

100

and passing through

(10, 500) : y = 100 x - 500

y = 100 x - 500

\frac{d ^{3}}{d x ^{3}} (tan (x))

y^{^{''}} + y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 2

\int - 8 x d x

\int \frac{x ^{4} cos ( x )}{8} d x

\int_{0}^{2} (4 x + 5) e^{- x} d x

What is the tangent of f(x)=5x^2,\at x=10 ?
The tangent of f(x)=5x^2,\at x=10 is y=100x-500