# Differentiation

#### cozza

Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

$$\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}$$

Rules: quotient rule, product rule, sum/multiple rules

forulma: $$\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}$$

$$\displaystyle e(x)=e^{tanx}$$, $$\displaystyle u=tanx$$
$$\displaystyle e'(u)=e^u$$, $$\displaystyle \frac{du}{dx}=sec^2x$$
$$\displaystyle \frac{de}{du}=e^u$$
$$\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}$$

$$\displaystyle g(x)=(x^3+1)^2$$, $$\displaystyle u=x^3+1$$
$$\displaystyle g'(u)=u^2$$, $$\displaystyle \frac{du}{dx}=3x^2$$
$$\displaystyle \frac{dg}{du}=2u$$
$$\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2$$

$$\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}$$
$$\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}$$

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)

#### Chris L T521

MHF Hall of Fame
Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

$$\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}$$

Rules: quotient rule, product rule, sum/multiple rules

forulma: $$\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}$$

$$\displaystyle e(x)=e^{tanx}$$, $$\displaystyle u=tanx$$
$$\displaystyle e'(u)=e^u$$, $$\displaystyle \frac{du}{dx}=sec^2x$$
$$\displaystyle \frac{de}{du}=e^u$$
$$\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}$$

$$\displaystyle g(x)=(x^3+1)^2$$, $$\displaystyle u=x^3+1$$
$$\displaystyle g'(u)=u^2$$, $$\displaystyle \frac{du}{dx}=3x^2$$
$$\displaystyle \frac{dg}{du}=2u$$
$$\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2$$

$$\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}$$
$$\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}$$

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)
I disagree with something here...if $$\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}$$, then by your method, $$\displaystyle e(x)=e^{\tan x}$$ and $$\displaystyle g(x)=x^3+1$$ (not $$\displaystyle (x^3+1)^2$$). As a result, you're quotient rule should give you $$\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}$$, which really can't be simplified further....

HallsofIvy

#### cozza

I disagree with something here...if $$\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}$$, then by your method, $$\displaystyle e(x)=e^{\tan x}$$ and $$\displaystyle g(x)=x^3+1$$ (not $$\displaystyle (x^3+1)^2$$). As a result, you're quotient rule should give you $$\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}$$, which really can't be simplified further....

Sorry, the original question shoul have said $$\displaystyle (x^3+1)^2$$, not $$\displaystyle (x^3+1)$$

Hi,

The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

$$\displaystyle f(x)=\frac{e^{tanx}}{(x^3+1)}$$

Rules: quotient rule, product rule, sum/multiple rules

forulma: $$\displaystyle f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}$$

$$\displaystyle e(x)=e^{tanx}$$, $$\displaystyle u=tanx$$
$$\displaystyle e'(u)=e^u$$, $$\displaystyle \frac{du}{dx}=sec^2x$$
$$\displaystyle \frac{de}{du}=e^u$$
$$\displaystyle \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}$$

where did the factor "x" come from?

$$\displaystyle g(x)=(x^3+1)^2$$, $$\displaystyle u=x^3+1$$
$$\displaystyle g'(u)=u^2$$, $$\displaystyle \frac{du}{dx}=3x^2$$
$$\displaystyle \frac{dg}{du}=2u$$
$$\displaystyle \frac{dg}{dx}=2(x^3+1)3x^2$$

$$\displaystyle f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}$$
$$\displaystyle f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}$$

I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?) (Worried)

$$\displaystyle f(x)=\displaystyle\huge\frac{e^{tanx}}{(x^3+1)^2}=e^{tanx}\frac{1}{(x^3+1)^2}=e^{tanx}\left(x^3+1\right)^{-2}$$

You can use the product rule...

$$\displaystyle f'(x)=\displaystyle\huge\frac{1}{(x^3+1)^2}\left(e^{tanx}sec^2x\right)+e^{tanx}\left(-2\frac{1}{(x^3+1)^3}3x^2\right)$$

then clean that up..

Last edited:
cozza