Answer to Question #151080 in Calculus for Angelo

Question

Differentiate the given trigonometric functions.

1. 𝑓(𝑥) = 3 𝑠𝑖𝑛2𝑥 𝑐𝑜𝑠4𝑥

2. 𝑦 = 1/2 𝑡𝑎𝑛^4(𝑠𝑒𝑐 6𝑥)

3. 𝑔(𝑡) = tan 𝑡/𝑐𝑜𝑠𝑡−4

4. 𝑦 = 8 𝑐𝑠𝑐 𝑥/4 𝑐𝑜𝑡 𝑥/4

5. 𝑓(𝑥) = 𝑥𝑠𝑖𝑛3𝑥 + 𝑥 𝑐𝑜𝑠3𝑥

Answer 1

1) Here $f(x)=3.sin2x.cos4x$

So, $f'(x)=\frac{d}{dx}(3.sin2x.cos4x)$

$=3.\frac{d}{dx}(sin3x.cos4x)$

$=3.[sin3x.\frac{d}{dx}(cos4x)+cos4x.\frac{d}{dx}(sin3x)]$

$=3.[sin3x.(-sin4x).\frac{d}{dx}(4x)+cos4x.cos3x.\frac{d}{dx}(3x)]$

$=3.[sin3x.(-sin4x).4+cos4x.cos3x.3]$

$=3.[-4.sin3x.sin4x+3.cos3x.cos4x]$

2) Here $y=(1/2)tan^4(sec6x)$

Now, $\frac{dy}{dx}=\frac{d}{dx}[\frac{1}{2}.tan^4(sec6x)]$

$=\frac{1}{2}.\frac{d}{dx}[tan^4(sec6x)]$

$=\frac{1}{2}.4.tan^3(sec6x).\frac{d}{dx}[tan(sec6x)]$

$=2.tan^3(sec6x).sec^2(sec6x).\frac{d}{dx}(sec6x)$

$=2.tan^3(sec6x).sec^2(sec6x).(sec6x.tan6x).\frac {d}{dx}(6x)$

$=2.tan^3(sec6x).sec^2(sec6x).sec6x.tan6x.6$

$=12.sec6x.tan6x.sec^2(sec6x).tan^2(sec6x)$

3) Here $g(t)=\frac{tant}{cost-4}$

Therefore, $g'(t)=\frac {d}{dt}[\frac{tant}{cost-4}]$

$=\frac{[(cost-4).\frac{d}{dt}(tant)]-[tant.\frac{d}{dt}(cost-4)]}{(cost-4)^2}$

$=\frac{[(cost-4).sec^2t]-[tant.(-sint)]}{(cost-4)^2}$

$=\frac{sec^2t.(cost-4)+tant.sint}{(cost-4)^2}$

4) Here $y=8.csc(\frac{x}{4}).cot(\frac{x}{4})$

Therefore, $\frac{dy}{dx}=\frac{d}{dx}[8.csc(\frac{x}{4}).cot(\frac{x}{4})]$

$=8[csc(\frac{x}{4}).\frac{d}{dx}(cot(\frac{x}{4}))+cot(\frac{x}{4}).\frac{d}{dx}(csc(\frac{x}{4}))]$

$=8[csc(\frac{x}{4}).(-csc^2(\frac{x}{4})).\frac{d}{dx}(\frac{x}{4})+cot(\frac{x}{4}).(-csc(\frac{x}{4}).cot(\frac{x}{4})).\frac{d}{dx}(\frac{x}{4})]$

$=8[-csc^3(\frac{x}{4}).\frac{1}{4}-csc(\frac{x}{4}).cot^2(\frac{x}{4}).\frac{1}{4}]$

$=-8.\frac{1}{4}.csc(\frac{x}{4})[csc^2(\frac{x}{4})+cot^2(\frac{x}{4})]$

$=-2.csc(\frac{x}{4})[csc^2(\frac{x}{4})+cot^2(\frac{x}{4})]$

5) Here $f(x)=x.sin3x+x.cos3x$

$=x.(sin3x+cos3x)$

Therefore, $f'(x)=\frac{d}{dx}[x.(sin3x+cos3x)]$

$=x.\frac{d}{dx}(sin3x+cos3x)+(sin3x+cos3x).\frac{d}{dx}(x)$

$=x.[cos3x.\frac{d}{dx}(3x)+(-sin3x).\frac{d}{dx}(3x)]+(sin3x+cos3x).1$

$=x.[3.cos3x-3.sin3x]+(sin3x+cos3x)$

$=(1-3x).sin3x+(1+3x).cos3x$