Calculus 1 Derivatives

[safe warren]

Good afternoon guys, I was wondering if anyone could explain to me and guide me with steps into solving these derivatives as halfway into solving the problem I would always get stuck.Your help would be very much appreciated

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[tawny bramble]

Do you know Product, Quotient and Chain rule

Product
Given $y=f(x)g(x)$, $\frac{dy}{dx}= f’(x)g(x) + f(x)g’(x)$

[wheat heath]

r•eⁱˣ = r•cos(x) + ri•sin(x)

[tawny bramble]

Chain:
Given y = f(g(x)), $\frac{dy}{dx} = f’(g(x))g’(x)$

[wheat heath]

r•eⁱˣ = r•cos(x) + ri•sin(x)

[safe warren]

yes I do know the rules

my issue is the step whereby i have to combine the product rule with the chain rule

[stone pewter]

tharts not hard

[trail viper]

You can use logarithmic differentiation for the first one it's easier

Take the natural logs of both sides and then differentiate w.r.t x

Lemme do the first step for you

$$y = \frac{2 \cos{x²}}{3x-1}$$

[wheat heath]

AstroVortex

[trail viper]

So after taking the natural logs

$$\ln{y} = \ln{\frac{2 \cos{x²}}{3x-1}}$$

[wheat heath]

AstroVortex

[trail viper]

Using prop6erties of logs:-

$$\ln{y} = \ln{2\cos{x²}} -\ln{(3x-1)}$$

[wheat heath]

AstroVortex

[trail viper]

You can further break it up if you want

$$\ln{y} = \ln{2} + 2\ln{\cos{x}} - \ln{(3x - 1)}$$

[wheat heath]

AstroVortex

[trail viper]

Now differentiate both sides w.r.t x

Ping me if you need further help

[trail viper]

@safe warren do you need further help?

[stone pewter]

@trail viper nah he doesn't

[trail viper]

Ig I'll solve them for him

with steps

The second one, y = sin(sin 2t) can be solved using the chain rule

According to chain rule, d/dx f(g(x)) = f'(g(x))•g'(x)

[trail viper]

In this case, let f(x) = sin x and g(x) = sin 2x

Now use the chain rule

So d/dx sin(sin 2t) = cos(sin 2t)•cos 2t

[safe warren]

Thank you so much guys ,sorry for the late reply

I have a better understanding now. I will come back to you if I have any more questions

[trail viper]

Yeah sure

Ping me if you need further help

[stone pewter]

he doesn't need further help bro

[half quiver]

Huh

Couldn’t u do some quotient rule

[trail viper]

yeah but i think logarithmic differentiation us easy there