Math Help - derivatives

1. derivatives

the derivative of

y = e^secx

is

y = e^secx correct?

2. Nope, not correct. You forgot to apply the chain rule.

Recall that:
$\frac{d}{dx} \sec{x} = \sec{x}\tan{x}$

3. Originally Posted by jvignacio
the derivative of

y = e^secx

is

y = e^secx correct?
Incorrect.

We know that:

$\frac{d}{dx}e^x=e^x$

But this time our exponent is $\sec{x}$, not $x$. So what we must do is substitute. Let $u=\sec{x}$. Then:

$e^{\sec{x}}=e^u$

$\frac{d}{du}e^u=e^u=e^{\sec{x}}$

$\frac{dy}{du}=e^{\sec{x}}$

But we can't stop there! What we have taken is $\frac{dy}{du}$. To get $\frac{dy}{dx}$, we have to multiply by $\frac{du}{dx}$, thusly:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

So, what is $\frac{du}{dx}$? Well, that's simple enough to find:

$u=\sec{x}$

$\frac{du}{dx}=\frac{d}{dx}\sec{x}=\sec{x}\tan{x}$

$\frac{du}{dx}=\sec{x}\tan{x}$

So:

$\frac{dy}{du}\frac{du}{dx}=e^{\sec{x}}\sec{x}\tan{ x}$

$\frac{dy}{dx}=e^{\sec{x}}\sec{x}\tan{x}$

In the future, you can skip writing down everything and just use this chain rule in your head.

4. Originally Posted by Chop Suey
Nope, not correct. You forgot to apply the chain rule.

Recall that:
$\frac{d}{dx} \sec{x} = \sec{x}\tan{x}$
y' = e^secx(secxtanx)

better?

5. Originally Posted by hatsoff
Incorrect.

We know that:

$\frac{d}{dx}e^x=e^x$

But this time our exponent is $\sec{x}$, not $x$. So what we must do is substitute. Let $u=\sec{x}$. Then:

$e^{\sec{x}}=e^u$

$\frac{d}{du}e^u=e^u=e^{\sec{x}}$

$\frac{dy}{du}=e^{\sec{x}}$

But we can't stop there! What we have taken is $\frac{dy}{du}$. To get $\frac{dy}{dx}$, we have to multiply by $\frac{du}{dx}$, thusly:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

So, what is $\frac{du}{dx}$? Well, that's simple enough to find:

$u=\sec{x}$

$\frac{du}{dx}=\frac{d}{dx}\sec{x}=\sec{x}\tan{x}$

$\frac{du}{dx}=\sec{x}\tan{x}$

So:

$\frac{dy}{du}\frac{du}{dx}=e^{\sec{x}}\sec{x}\tan{ x}$

$\frac{dy}{dx}=e^{\sec{x}}\sec{x}\tan{x}$