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junior college 2 | H1 Maths
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QN
QN

junior college 2 chevron_right H1 Maths chevron_right Singapore

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Date Posted: 3 years ago
Views: 380
J
J
3 years ago
d/dx e^(x²+2)

= ( d/dx (x² + 2) ) e^(x²+2)

= 2x e^(x² + 2)
J
J
3 years ago
Alternative method : implicit differentiation


let y = e^(x²+2)

Take ln on both sides,

ln y = ln e^(x² + 2)

ln y = x² + 2

Differentiate both sides w.r.t x,

1/y dy/dx = 2x

dy/dx = 2xy = 2xe^(x² + 2)
QN
QN
3 years ago
I don’t get the first one...
If I differentiate exponentials shouldnt it be d/dx(e^f(x)) = f’(x)e^f(x)
J
J
3 years ago
That's basically what I did.
J
J
3 years ago
I wrote (d/dx (x² + 2)) e^(x² + 2)

d/dx f(x) is the same as f'(x)
QN
QN
3 years ago
Oh I see haha thanks a lot!
J
J
3 years ago
Exactly
QN
QN
3 years ago
Sry I didn’t notice when I first saw it

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Ignatius Yeo
Ignatius Yeo's answer
36 answers (Tutor Details)
1st
The picture shows 2 methods of doing it - product rule and chain rule. For chain rule, let f(x) be x^2+2. Differentiate f(x) and multiply e^f(x) as shown. Hope it helps!
QN
QN
3 years ago
Thanks, I get the second one!
But the first one, why is e^x^2 when differentiated is e^x^2(0)?
Ignatius Yeo
Ignatius Yeo
3 years ago
Whenever you differentiate a constant value, it is 0. The first method is product rule. First term multiply by derivative of second term, and second term multiply derivative of first term. e^2 is a constant with no variables. Therefore d/dx e^2=0.
J
J
3 years ago
Product rule...

d/dx (e^(x²) e²)

= e^(x²) d/dx (e²) + (d/dx e^(x²) ) e²

= e^(x²) (0) + 2xe^(x²) (e²)

= 2xe²e^(x²)

= 2x e^(x²+2)


Remember that differentiating a constant gives 0
Ignatius Yeo
Ignatius Yeo
3 years ago
Another tip d/dx e^2 = d/dx e^2(x^0). Applying differentiation rule, d/dx (e^2(x^0))= 0*x^-1*e^2=0. Anything multiplies 0 is always 0. Hence, d/dx constant =0.
QN
QN
3 years ago
Thanks a lot!!
Ignatius Yeo
Ignatius Yeo
3 years ago
You're welcome.