# Differentiation and finding tangent

## Homework Statement

Find the equation of the tangent to the curve y = x2(x + 1)4 at the point P(1,16)

## The Attempt at a Solution

dy/dx x2(x + 1)4

= (x + 1)3((x + 1)2x + 4x2)

= (x + 1)3(6x2 + 2x)

= (x + 1)3(2x)(3x + 1)

Subst. 1 in to find grad.

(1 + 1)3(2)(3 + 1)

= 64

Seems wrong..

y - 16 = 64(x - 1)

y = 64x - 48

I think it's probably wrong, just so confused - can someone give me a hand please? Much appreciated ## Homework Statement

Find the equation of the tangent to the curve y = x2(x + 1)4 at the point P(1,16)

## The Attempt at a Solution

dy/dx x2(x + 1)4

= (x + 1)3((x + 1)2x + 4x2)

= (x + 1)3(6x2 + 2x)

= (x + 1)3(2x)(3x + 1)

Subst. 1 in to find grad.

(1 + 1)3(2)(3 + 1)

= 64

Seems wrong..

y - 16 = 64(x - 1)

y = 64x - 48

I think it's probably wrong, just so confused - can someone give me a hand please? Much appreciated You should plug it into the derivative formula

$$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$

and put this into you Calculator...

and use the good old Tangent formula.

$$y = f(a) + f'(a)(x-a)$$ where you take the derivative at x = a.

Thank you for your reply, Susanne, it's just that I've never seen the tangent formula before :S

Also my homework is based around the idea of differentiating so I was wondering if I was going in the right direction :S

Thank you for your reply, Susanne, it's just that I've never seen the tangent formula before :S

Also my homework is based around the idea of differentiating so I was wondering if I was going in the right direction :S

You derivative is correct.

But you can use the definition of the derivative and graphical Calculator to check your result.

and your userage of the tangent formula looks to be okay too :)

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Oo okay, thank you - just thought it was wrong considering the grad was 64 that's all.

Thanks Susanne for checking :D