tangent of y=x^3+x — step-by-step solution

No x-coordinate is specified: find the tangent at the general point

x = a

, for real

a

Let's differentiate:

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

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

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

ⓘ The derivative of x^n is n*x^(n-1):

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

3 x^{2}

3 x^{2} + 1

Compute f(

a

)

The point of tangency is (

a

a^{3} + a

)

Compute f'(

a

)

The slope is f'(

a

) =

3 a^{2} + 1

The tangent passes through this point with this slope:

y = (3 a^{2} + 1) \cdot (x - a) + (a^{3} + a)

Compute the y-intercept:

b = (a^{3} + a) - (3 a^{2} + 1) \cdot a

(a^{3} + a) - (3 a^{2} + 1) \cdot a

a^{3} + a - (3 a^{2} + 1) \cdot a

ⓘ Expand by multiplying each term inside the parentheses:

- (3 a^{2} + 1) \cdot a

- (3 a^{3} + a)

a^{3} + a - (3 a^{3} + a)

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (3 a^{3} + a)

- 3 a^{3} - a

a^{3}

+ a

- 3 a^{3}

- a

ⓘ Like terms are grouped by adding their coefficients:

a^{3} - 3 a^{3}

-2 a^{3}

a - a

0

- 2 a^{3} + 0

ⓘ Adding 0 changes nothing:

- 2 a^{3} + 0

-2 a^{3}

- 2 a^{3}

▶Tangent line:

y = (3 a^{2} + 1) \cdot x - 2 a^{3}