Quick questions on Differentiation of inverse circular functions: VCE Specialist Mathematics Unit 4

Q: What is first function?

Here the inner function is u = 3x, so u' = 3. Applying (d)/(dx)(u) = (u')/(1 + u^2):

Q: What is second function?

Here u = x^2, so u' = 2x. Applying (d)/(dx)(u) = (u')/(sqrt(1 - u^2)):

Q: What is mismatched a?

In the arctan antiderivative the factor is (1)/(a), and the argument is (x)/(a). Keep them consistent.

6short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is first function?

Show answer

Here the inner function is

u = 3x

, so

u' = 3

. Applying

\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}

What is second function?

Show answer

Here

u = x^2

, so

u' = 2x

. Applying

\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}

What is mismatched

a

Show answer

In the arctan antiderivative the factor is

\frac{1}{a}

, and the argument is

\frac{x}{a}

. Keep them consistent.

What is q1?

Show answer

Differentiate

y = \arcsin x

. [1 mark]

What is q2?

Show answer

Differentiate

y = \arctan(x^3)

. [2 marks]

What is q3?

Show answer

Evaluate

\int\frac{dx}{\sqrt{25 - x^2}}

. [2 marks]

Have a question we have not covered?

This dot-point answer is short enough that we have not extracted many short questions yet. Read the full dot-point answer or ask Mo, our study assistant, in the chat for follow ups.

All Specialist MathematicsQ&A pages