Find the derivative
1. y = e^xSin²x
2. y = (x³+1)e^-4
3. y = x²e^Sinx
if y= uv
y'= u v' + v u'
int the first, u=e^x v=sin^2x
du= e^x dx dv=2sinxcosx dx
y= e^x *2sinx cosx + e^x sin^2x
I will be happy to critique your thinking on the others.
Nevermind, please don't answer this question. I have made a mistake. Thanks.
Oh but thank you.
Ok, I wrote number 2 wrong. Number 1 and number 3 equations are correct but number 2 is actually y = (x²+1)e^-4
1) y = e^x(Sin²x)
2) y = (x²+1)e^-4
3) y = x²(e^Sinx)
please help me on all 3 questions. thank you.
To find the derivative of a given function, we will use the rules of differentiation.
1. y = e^xSin²x
To differentiate this function, we will apply the product rule. The product rule states that if we have two functions, u(x) and v(x), their derivative can be found using the formula: (u*v)' = u'v + uv'.
u(x) = e^x
v(x) = sin²x
First, we need to find the derivative of u(x) and v(x):
u'(x) = d(e^x)/dx = e^x
v'(x) = d(sin²x)/dx = 2sin(x)cos(x) (using the chain rule)
Now, we can apply the product rule:
y' = u'v + uv' = e^x*sin²x + e^x*2sin(x)cos(x)
Simplifying further:
y' = e^x(sin²x + 2sin(x)cos(x))
2. y = (x³+1)e^-4
To differentiate this function, we will apply the product rule again.
u(x) = (x³+1)
v(x) = e^-4
First, we find the derivatives of u(x) and v(x):
u'(x) = d(x³+1)/dx = 3x²
v'(x) = d(e^-4)/dx = -4e^-4 (since the derivative of e^ax with respect to x is ae^ax)
Applying the product rule:
y' = u'v + uv' = (3x²)(e^-4) + (x³+1)(-4e^-4)
Simplifying further:
y' = 3x²e^-4 - 4(x³+1)e^-4
3. y = x²e^sinx
To find the derivative of this function, we will apply the product rule once again.
u(x) = x²
v(x) = e^sinx
Finding the derivatives:
u'(x) = d(x²)/dx = 2x
v'(x) = d(e^sinx)/dx = cosx * e^sinx (using the chain rule)
Applying the product rule:
y' = u'v + uv' = (2x)(e^sinx) + (x²)(cosx * e^sinx)
Simplifying further:
y' = 2xe^sinx + x²cosx * e^sinx
And that's how you find the derivatives of the given functions.