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

The tangent of f(x)=x^5,\at x=1 is y=5x-4

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

tangent of

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

y = 5 x - 4

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = x^{5} : \frac{df ( x )}{d x} = 5 x^{4}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5

Find the line with slope m=

5

and passing through

(1, 1) : y = 5 x - 4

y = 5 x - 4

\frac{d}{d x} (\frac{3 x}{x ^{3} + 1})

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

\int \frac{sin ^{5} ( x )}{cos ^{2} ( x )} d x

\frac{\partial}{\partial x} (6 y e^{7 x})

\frac{d y}{d x} = (\frac{x}{y - 2})^{{3}} ln (2 x)

What is the tangent of f(x)=x^5,\at x=1 ?
The tangent of f(x)=x^5,\at x=1 is y=5x-4