What is the tangent of f(x)=sqrt(x)(x^6-1),\at x=1 ?

The tangent of f(x)=sqrt(x)(x^6-1),\at x=1 is y=6x-6

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tangent of f(x)=sqrt(x)(x^6-1),\at x=1

tangent of

f (x) = x (x^{6} - 1), at x = 1

y = 6 x - 6

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x (x^{6} - 1) : \frac{df ( x )}{d x} = \frac{13 x ^{6} - 1}{2 x}

Compute the slope of

f (x) = x (x^{6} - 1)

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 0) : y = 6 x - 6

y = 6 x - 6

\int e^{x} \cdot cos (3 x) d x

\int \frac{6 x ^{2} - 15 x + 22}{( x + 3 ) ( x ^{2} + 2 ) ^{2}} d x

y = \frac{1}{x}, (4, \frac{1}{2})

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 t}

\frac{\partial}{\partial x} (x^{s i n (x)})

What is the tangent of f(x)=sqrt(x)(x^6-1),\at x=1 ?
The tangent of f(x)=sqrt(x)(x^6-1),\at x=1 is y=6x-6