Find the dervatives:
1. f(x)=(3x+1)e^x^2
2. y = e^(sin x) ln(x)
3. f(x)=(x^(2)+x)^23
4. f(x)=sin(2x)/cosx
5. square root of x/(3x+1)
6. f(x)=sin^4(3x=1)-sin(3x+1)
7. x+y=cos(xy)
The answers I got:
1. 6x^(2)e^(x^2)+2xe^(x^2)+3e^(x^2)
2. e^sinx/x + -lnxe^(sinx)cosx
3. 46x+23(x^2+x)^22
4. ?
5. 3x+1-3squarerootofx/(2squarerootofx)(3x+1)^w
6. ?
7. (-ysin(xy)-1)/(2+xsin(xy))
1. messy, but correct. I'd have gone on to
(6x^2 + 2x + 3)e^(x^2)
2. almost. Why the "-" sign? d(sinx)/dx = +cosx
e^(sinx) (1/x + lnx cosx)
3. correct.
23(2x+1)(x^2+x)^22
4. brute force, using the quotient rule:
((2cos 2x)(cosx) - sin 2x (-sinx))/cosx)^2
= (2 cosx cos2x + sinx sin2x)/(cosx)^2
messy. How about simplifying first?
sin2x/cosx = 2sinx cosx / cosx = 2sinx
f' = 2cosx
I'll let you convince yourself that the two are equal
5. Hmmm. I get
f = √x/(3x+1)
f' = ((1/2√x)(3x+1) - √x(3))/(3x+1)^2
= ((3x+1) - 3√x 2√x)/(2√x (3x+1)^2)
= (1-3x)/(2√x (3x+1)^2)
a little algebra mixup in the top?
6. If you mean
f = sin^4(3x+1)-sin(3x+1), just use the good old chain rule on both terms:
f' = 4sin^3(3x+1)(cos(3x+1))(3) - cos(3x+1)(3)
= 3cos(3x+1)(4sin^3(3x+1) - 1)
7. where'd that 2 come from?
1 + y' = -sin(xy)(y+xy')
1 + y' = -ysin(xy) - xsin(xy) y'
y' = -(1+y sin(xy))/(1+x sin(xy))
Dervative of cos(x) is -sin(x) so it would be negative.
To find the derivatives of the given functions, we need to apply the appropriate rules. Let's go through each question step by step:
1. f(x) = (3x + 1)e^(x^2)
To find the derivative, we can use the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by u'(x)v(x) + u(x)v'(x).
Let's assign u(x) = 3x + 1 and v(x) = e^(x^2).
1a. Find u'(x):
The derivative of u(x) = 3x + 1 with respect to x is u'(x) = 3.
1b. Find v'(x):
To find the derivative of v(x), which is e^(x^2), we can use the chain rule. The chain rule states that if we have a function f(g(x)), then the derivative of f(g(x)) with respect to x is f'(g(x)) * g'(x).
In this case, let's assign f(u) = e^u and g(x) = x^2.
1b(i). Find f'(u):
The derivative of f(u) = e^u with respect to u is f'(u) = e^u.
1b(ii). Find g'(x):
The derivative of g(x) = x^2 with respect to x is g'(x) = 2x.
1b(iii). Combine f'(u) and g'(x):
Using the chain rule, v'(x) = f'(u) * g'(x) = e^(x^2) * 2x.
1c. Combine u'(x) and v'(x):
Now that we have u'(x) = 3 and v'(x) = e^(x^2) * 2x, we can use the product rule to find the derivative of f(x):
f'(x) = u'(x) * v(x) + u(x) * v'(x)
= 3 * e^(x^2) + (3x + 1) * e^(x^2) * 2x
= 3e^(x^2) + (6x^2 + 2x)e^(x^2)
= 6x^2e^(x^2) + 2xe^(x^2) + 3e^(x^2)
So the derivative of f(x) with respect to x is 6x^2e^(x^2) + 2xe^(x^2) + 3e^(x^2).
Let's move on to the next question.