- #1

Emjay

- 4

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f(x)=(2cos^2 x+3)^5/2

Any help would be very much appreciated:)

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- Thread starter Emjay
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In summary, the conversation is about someone seeking help with finding the derivative of a function involving cosine. The expert explains the power and chain rules and offers guidance on how to proceed with the differentiation. The person expresses their gratitude and mentions they have another question to ask.

- #1

Emjay

- 4

- 0

f(x)=(2cos^2 x+3)^5/2

Any help would be very much appreciated:)

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- #2

MarkFL

Gold Member

MHB

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Just to be clear, we are given:

\(\displaystyle f(x)=\left(2\cos^2(x)+3\right)^{\Large{\frac{5}{2}}}\)

And we are asked to compute $f'(x)$...correct?

If that's not correct, please let us know, but if it is...can you post what you've tried? This way we can see what you might be doing wrong. :)

- #3

Emjay

- 4

- 0

Honestly, not even sure where to start with this one.

I have only ever done basic examples and this one has got me stumped.

- #4

MarkFL

Gold Member

MHB

- 13,288

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Emjay said:

Honestly, not even sure where to start with this one.

I have only ever done basic examples and this one has got me stumped.

Okay, suppose we have:

\(\displaystyle g(x)=\left(h(x)\right)^r\)

The power and chain rules tell us:

\(\displaystyle g'(x)=r\left(h(x)\right)^{r-1}h'(x)\)

Using this formula, we can then write:

\(\displaystyle f'(x)=\frac{5}{2}\left(2\cos^2(x)+3\right)^{\Large{\frac{3}{2}}}\left(2\cos^2(x)+3\right)'\)

Can you proceed, using the formula again to compute the indicated differentiation?

- #5

Emjay

- 4

- 0

That helped heaps, thank you.

I do have a separate question to ask so i'll re post

Thanks again :)

I do have a separate question to ask so i'll re post

Thanks again :)

The basic trigonometric functions are sine, cosine, and tangent. They are commonly denoted as sin(x), cos(x), and tan(x) respectively.

To find the derivative of a trigonometric function, you can use the following rules:

- The derivative of sine is cosine (d/dx(sin(x)) = cos(x))

- The derivative of cosine is negative sine (d/dx(cos(x)) = -sin(x))

- The derivative of tangent is secant squared (d/dx(tan(x)) = sec^2(x))

Yes, you can use the chain rule to find the derivative of a trigonometric equation. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). This can be applied to trigonometric equations by substituting in the appropriate derivative for the inner function.

Yes, there are a few special cases when finding the derivative of a trigonometric function. For example, the derivative of secant is secant*tangent (d/dx(sec(x)) = sec(x)*tan(x)). Also, the derivative of cotangent is negative cosecant squared (d/dx(cot(x)) = -csc^2(x)).

To find the derivative of a trigonometric equation with multiple trigonometric functions, you can use the product rule or quotient rule depending on the form of the equation. The product rule states that if y = f(x)*g(x), then dy/dx = f'(x)*g(x) + f(x)*g'(x). The quotient rule states that if y = f(x)/g(x), then dy/dx = (f'(x)*g(x) - f(x)*g'(x))/(g(x))^2.

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