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junior college 2 | H1 Maths
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Date Posted: 4 years ago
Views: 426
d/dx e^(x²+2)
= ( d/dx (x² + 2) ) e^(x²+2)
= 2x e^(x² + 2)
= ( d/dx (x² + 2) ) e^(x²+2)
= 2x e^(x² + 2)
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)
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)
I don’t get the first one...
If I differentiate exponentials shouldnt it be d/dx(e^f(x)) = f’(x)e^f(x)
If I differentiate exponentials shouldnt it be d/dx(e^f(x)) = f’(x)e^f(x)
That's basically what I did.
I wrote (d/dx (x² + 2)) e^(x² + 2)
d/dx f(x) is the same as f'(x)
d/dx f(x) is the same as f'(x)
Oh I see haha thanks a lot!
Exactly
Sry I didn’t notice when I first saw it
See 1 Answer
done
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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!
Date Posted:
4 years ago
Thanks, I get the second one!
But the first one, why is e^x^2 when differentiated is e^x^2(0)?
But the first one, why is e^x^2 when differentiated is e^x^2(0)?
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.
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
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
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.
Thanks a lot!!
You're welcome.
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