# How do you solve these questions. As i keep getting them wrong (1 Viewer)

#### synthesisFR

##### Well-Known Member
what are you struggling with? understanding chain rule or something else?

#### Average Boreduser

##### “If things go wrong, don’t go with them.”- Roger B
is this prelim or hsc 2U?

#### mmmmmmmmaaaaaaa

##### Well-Known Member
What exactly are you struggling with in these questions? Which step of the question do you get stuck on

#### cossine

##### Well-Known Member
Theorems
sin x(d/dx)= cos x

cos x(d/dx) = -sin x

-sin x(d/dx) = -cos x

-cos x(d/dx) = sin x

The proof of these theorems is outside the scope of high school however nothing will stop you from applying the theorem.

Other than that question are fairly straightforward, so you may need to apply product rule or chain rule

#### cossine

##### Well-Known Member
Just remember these:

y = sin [ f(x) ]
y' = f'(x) x cos [ f(x) ]

y = cos [ f(x) ]
y' = f'(x) x -sin [ f(x) ]

y = tan [ f(x) ]
y' = f'(x) x sec^2 [ f(x) ]

Let's apply this to the question above:

y = sin [ 1/4x + pi/2 ]

So f(x) = 1/4x + pi/2

f'(x) = 1/4

Therefore y' = 1/4 x cos [ 1/4x + pi/2 ]

But the question is asking for y'(pi)

So in place of x, we now substitute x = pi

Doing this we get y'(pi) = 1/4 x cos [ 1/4(pi) + pi/2 ] = ________

You can take it step by step until you reach your answer, don't get overwhelmed
You could do this but it does not demonstrate understanding of chain rule.

As a side note you should use * for multiplication has "x" can be confused with the variable "x".

#### Dane Red

##### Member
The problem is that the answer i get is different to the answer in the text book

#### tywebb

##### dangerman
So this is year 12 Cambridge Advanced 6B Q1u, 4d, 5d, 12a

It does seem that the answer in the textbook for this one is wrong. Textbook answer is $\bg_white 4\cos\frac{x}{4}$ but it should be $\bg_white 3\cos\frac{x}{4}$

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