We do the usual way we learnt in Sec 4.

Oh yes. Do note that sin (x⁴) is NOT the same as sin⁴x.

y = x³ sin x⁴.

Using product rule and chain rule,

dy/dx
= x³ times d/dx (sin x⁴) + (sin x⁴) times d/dx (x³)
= x³ times (cos x⁴) times d/dx (x⁴) + (sin x⁴) times 3x²
= x³ times cos x⁴ times 4x³ + 3x² times sin x⁴
= 4x⁶ times cos x⁴ + 3x² times sin x⁴

When x = 1,
dy/dx
= 4 times cos 1 + 3 times sin 1
= 4.685622178
~ 4.69
[angles are in radian mode]

We have the gradient as 4.685622178. This will be the gradient of the tangent as well (since the gradient of a curve at a point is taken to be the gradient of the tangent drawn through that point).

As the case is in Sec 4, we need a coordinate to start the equation of a tangent (or any line in general).

When x = 1,
y = 1 sin 1

I will write the gradient as (4 cos 1 + 3 sin 1).

y = mx + c
y = (4 cos 1 + 3 sin 1) x + c

We sub in x and y.

1 sin 1 = (4 cos 1 + 3 sin 1) (1) + c

Solving for c, c = -4 cos 1 - 2 sin 1.

y = (4 cos 1 + 3 sin 1) x - 4 cos 1 - 2 sin 1
y = 4.685622178x - 3.844151193
y = 4.69x - 3.84

Similar to the previous examples, observe that 4x³ is the derivative of x⁴.

This part is settled.

What does integrating sine become? A cosine. Or more accurately, a negative cosine.

INT [x³ sin x⁴] dx from 0 to b
= 1/4 times INT [4x³ sin x⁴] from 0 to b
= 1/4 times [-cos x⁴] from 0 to b

[The 4x³ is part of the chain and gets absorbed into the integral, similar to the integral you did the other day except that here it's for trigonometric functions]

= 1/4 times [-cos 0 + cos b⁴]
= 1/4 times [cos b⁴ - 1]

thanks! i understand it now. i also have another question regarding calculus too which i just posted, is it possible to to help me with the question too?