What is the tangent of f(x)=5x^3-5x,\at x=1 ?

The tangent of f(x)=5x^3-5x,\at x=1 is y=10x-10

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tangent of f(x)=5x^3-5x,\at x=1

tangent of

f (x) = 5 x^{3} - 5 x, at x = 1

y = 10 x - 10

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 5 x^{3} - 5 x : \frac{df ( x )}{d x} = 15 x^{2} - 5

Compute the slope of

f (x) = 5 x^{3} - 5 x

x = 1 : m = 10

Find the line with slope m=

10

and passing through

(1, 0) : y = 10 x - 10

y = 10 x - 10

\frac{d}{d x} (e^{3 - x^{2}})

n = 4 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

\int \frac{x}{( 2 - x ^{2} ) ^{2}} d x

y^{^{'}} = \frac{( 6 x ^{4} + y ^{4} )}{x y ^{3}}

\int_{0}^{\frac{π}{2}} 18 cos^{5} (x) d x

What is the tangent of f(x)=5x^3-5x,\at x=1 ?
The tangent of f(x)=5x^3-5x,\at x=1 is y=10x-10