1. ## Re: Differentiate the functions.

The first one is correct. The second is not, the first term of the second derivative should actually be \displaystyle \begin{align*} -\frac{3}{x^4} \end{align*}.

This is because \displaystyle \begin{align*} \frac{1}{x^3} = x^{-3} \end{align*}, which you would then use the power rule for.

2. ## Re: Differentiate the functions.

Thanks guys for your help! I really appreciate it! Could you tell me if I did these four problems correctly?

3. ## Re: Differentiate the functions.

c) Your first two terms are correct. For the third term, rewrite it as:

$\log_5(e^x)=x\log_5(e)$

Now you just have a constant times x.

For the 4th term, let's write:

$y=2^{\sqrt{x}}$

take the natural log of both sides:

$\ln(y)=\sqrt{x}\ln(2)$

Implicitly differentiate:

$\frac{1}{y}\cdot\frac{dy}{dx}=\frac{\ln(2)}{2\sqrt {x}}$

$\frac{dy}{dx}=y\frac{\ln(2)}{2\sqrt{x}}$

$\frac{dy}{dx}=2^{\sqrt{x}}\frac{\ln(2)}{2\sqrt{x}}$

d) For the first term you need to apply the product rule, and for the second term use:

$\frac{d}{dx}\left(\tan^{-1}(u(x)) \right)=\frac{1}{u^2+1}\cdot\frac{du}{dx}$

e) Apply the quotient rule.

f) Correct.

4. ## Re: Differentiate the functions.

Originally Posted by MarkFL2
c) Your first two terms are correct. For the third term, rewrite it as:

$\log_5(e^x)=x\log_5(e)$

Now you just have a constant times x.

For the 4th term, let's write:

$y=2^{\sqrt{x}}$

take the natural log of both sides:

$\ln(y)=\sqrt{x}\ln(2)$

Implicitly differentiate:

$\frac{1}{y}\cdot\frac{dy}{dx}=\frac{\ln(2)}{2\sqrt {x}}$

$\frac{dy}{dx}=y\frac{\ln(2)}{2\sqrt{x}}$

$\frac{dy}{dx}=2^{\sqrt{x}}\frac{\ln(2)}{2\sqrt{x}}$

d) For the first term you need to apply the product rule, and for the second term use:

$\frac{d}{dx}\left(\tan^{-1}(u(x)) \right)=\frac{1}{u^2+1}\cdot\frac{du}{dx}$

e) Apply the quotient rule.

f) Correct.
If you have not learnt implicit differentiation, you can do this...

\displaystyle \begin{align*} 2^{\sqrt{x}} &= 2^{x^{\frac{1}{2}}} \\ &= e^{\ln{\left( 2^{x^{\frac{1}{2}}} \right)}} \\ &= e^{x^{\frac{1}{2}}\ln{2}} \end{align*}

and now you can use the Chain Rule.

5. ## Re: Differentiate the functions.

I will get to those problems above later since I need to finish these other problems for tomorrow because they are due.
Are these two correct?

6. ## Re: Differentiate the functions.

h) When applying the prodcut rule to the first term, you neglected to apply the chain rule to the tangent function.

i) You differentiated correctly. I would choose to write the final result as:

$f'(x)=-3(x^2+1)\sin(2x^3+3x)\cos(\cos(2x^3+3x))$

7. ## Re: Differentiate the functions.

Thanks. How about these last two?

8. ## Re: Differentiate the functions.

g) You neglected to apply the chain rule to the argument of the trigonometric term in the numerator. Otherwise, it is correct.

j) In the denominator of the result, you want the first binomial factor to be squared. I would write the result as:

$f'(x)=\frac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}$

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