If
y = f(x) = e^x
then find:
1) f'(x) or dy/dx
2)f''(x) or d²y/dx²
Show your calculations as well
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NOTE: If you are thinking you can cancel d in the fractions.... NO, you can't. This is finding derivative. Maybe you just don't know what it is
Answers
Answer:
Answer:
Step-by-step explanation:
y=x^{x}y=x
x
Taking log on both sides,
logy=xlogxlogy=xlogx
Differentiating with respect to x, we have
\frac{1}{y}(\frac{dy}{dx})=x{\times}\frac{1}{x}+logx
y
1
(
dx
dy
)=x×
x
1
+logx
\frac{dy}{dx}=y+ylogx
dx
dy
=y+ylogx
\frac{dy}{dx}=y(1+logx)
dx
dy
=y(1+logx)
Again differentiating with respect to x,
\frac{d^{2}y}{dx^{2}}=y{\times}\frac{1}{x}+(1+logx)\frac{dy}{dx}
dx
2
d
2
y
=y×
x
1
+(1+logx)
dx
dy
\frac{d^{2}y}{dx^{2}}=\frac{y}{x}+y(1+logx)^{2}
dx
2
d
2
y
=
x
y
+y(1+logx)
2
Now, the given equation is:
\frac{d^{2}y}{dx^{2}}-\frac{1}{y}(\frac{dy}{dx})^{2}-\frac{y}{x}
dx
2
d
2
y
−
y
1
(
dx
dy
)
2
−
x
y
=\frac{y}{x}+y(1+logx)^{2}- y(1+logx)^{2}-\frac{y}{x}
x
y
+y(1+logx)
2
−y(1+logx)
2
−
x
y
=00
Answer:
Since l∥m, then
3y=2y+25
0
[Alternate ∠s]
⇒3y−2y=25
0
⇒y=25
0
..(1)
Also x+15
0
=2y+25
0
[Vertically opposite ∠s]
⇒x+15
0
=2(25)+25
0
[using (1)]
⇒x+15
0
=50
0
+25
0
⇒x=75
0
−15
0
⇒x=60
0
.(in degrees)