tangent of f(x)=(1+4x)^9,(0,1) — step-by-step solution

Find the tangent at

x = 0

Let's differentiate:

\frac{d}{d x} ({(1 + 4 x)}^{9})

We differentiate using:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

(where f =

1 + 4 x

)

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} ({(1 + 4 x)}^{9})

9 \cdot \frac{d}{d x} (1 + 4 x) \cdot {(1 + 4 x)}^{8}

9 \cdot \frac{d}{d x} (1 + 4 x) \cdot {(1 + 4 x)}^{8}

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

\frac{d}{d x} (1 + 4 x)

0 + \frac{d}{d x} (4 x)

9 \cdot (0 + \frac{d}{d x} (4 x)) \cdot {(1 + 4 x)}^{8}

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (4 x)

\frac{d}{d x} (4 x)

■A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (4 x)

4

9 \cdot 4 \cdot {(1 + 4 x)}^{8}

ⅰ

9 \cdot 4

36

36 \cdot {(1 + 4 x)}^{8}

36 \cdot {(4 x + 1)}^{8}

Compute f(

0

)

{(1 + 4 \cdot 0)}^{9}

ⓘ A product with a zero factor is zero:

4 \cdot 0

0

{(1 + 0)}^{9}

ⓘ Adding 0 changes nothing:

1 + 0

1

1

The point of tangency is (

0

1

)

Compute f'(

0

)

36 \cdot {(4 \cdot 0 + 1)}^{8}

ⓘ A product with a zero factor is zero:

4 \cdot 0

0

36 \cdot {(0 + 1)}^{8}

ⓘ Adding 0 changes nothing:

0 + 1

1

36

The slope is f'(

0

) =

36

The tangent passes through this point with this slope:

y = 36 x + 1

Compute the y-intercept:

b = 1 - 36 \cdot 0

1 - 36 \cdot 0

ⓘ A product with a zero factor is zero:

36 \cdot 0

0

1 + 0

ⓘ Adding 0 changes nothing:

1 + 0

1

1

▶Tangent line:

y = 36 x + 1