Differentiating trig equations

Hi, can anyone help me with the two questions circled?
In the box is the answer that I'm getting - which is wrong, so can anyone briefly lead me through what the correct method to solving it would be? Thanks! :smile:

Reply 1

TenOfThem

18

Original post by clarewithnoi

Hi, can anyone help me with the two questions circled?
In the box is the answer that I'm getting - which is wrong, so can anyone briefly lead me through what the correct method to solving it would be? Thanks! :smile:

xSec^2x \not= Sec^2x^2

Reply 2

TenOfThem

18

Original post by clarewithnoi

Hi, can anyone help me with the two questions circled?
In the box is the answer that I'm getting - which is wrong, so can anyone briefly lead me through what the correct method to solving it would be? Thanks! :smile:

I cannot even begin to imagine what your thinking was on the

\dfrac{\cos x}{x}

Reply 3

Kvothe the Arcane

20

Original post by clarewithnoi

Hi, can anyone help me with the two questions circled?
In the box is the answer that I'm getting - which is wrong, so can anyone briefly lead me through what the correct method to solving it would be? Thanks! :smile:

For the first box, use the quotient rule, 2nd the product rule.

Reply 4

L'Evil Fish

18

y = cos(x)/x

Quotient rule:

y = (x)(-sin(x)) - cos(x) / x^2

Reply 5

BabyMaths

\frac{d(x \tan x)}{dx} = \frac{d (x)}{dx} \tan x + x \frac{d(\tan x)}{dx}= \tan x + x \sec ^2 x

Reply 6

clarewithnoi

OP

Original post by TenOfThem

I cannot even begin to imagine what your thinking was on the

\dfrac{\cos x}{x}

I did the exact same haha but tried to take 'x' out as a common factor, thanks so much for the help - I missed this whole topic so I'd been trailing my hair out over it this morning. Just re-did them and now have the correct answer. Thanks so much! :smile:

Reply 7

clarewithnoi

OP

Original post by BabyMaths