Which Co-ordinate axis will help to move the sprite in Pictoblox left and right? *
Answers
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given expression is
\begin{gathered}\rm \: {x}^{a} \: {y}^{b} = {(x + y)}^{a + b} \\ \end{gathered}
x
a
y
b
=(x+y)
a+b
On taking log on both sides, we get
\begin{gathered}\rm \: log[{x}^{a} \: {y}^{b} ]= log{(x + y)}^{a + b} \\ \end{gathered}
log[x
a
y
b
]=log(x+y)
a+b
We know,
\begin{gathered}\boxed{\sf{ \:log {x}^{y} = ylogx \: }} \\ \end{gathered}
logx
y
=ylogx
and
\begin{gathered}\boxed{\sf{ \:log(xy) = logx + logy \: }} \\ \end{gathered}
log(xy)=logx+logy
So, using these results, we get
\begin{gathered}\rm \: alogx + blogy = (a + b)log(x + y) \\ \end{gathered}
alogx+blogy=(a+b)log(x+y)
On differentiating both sides w. r. t. x, we get
\begin{gathered}\rm \:\dfrac{d}{dx}\bigg( alogx + blogy\bigg) =\dfrac{d}{dx} (a + b)log(x + y) \\ \end{gathered}
dx
d
(alogx+blogy)=
dx
d
(a+b)log(x+y)
We know,
\begin{gathered}\boxed{\sf{ \:\dfrac{d}{dx}logx = \frac{1}{x} \: }} \\ \end{gathered}
dx
d
logx=
x
1
So, using this result, we get
\begin{gathered}\rm \: \dfrac{a}{x} + \dfrac{b}{y}\dfrac{dy}{dx}=(a + b) \dfrac{1}{x + y}\dfrac{d}{dx}(x + y) \\ \end{gathered}
x
a
+
y
b
dx
dy
=(a+b)
x+y
1
dx
d
(x+y)
\begin{gathered}\rm \: \dfrac{a}{x} + \dfrac{b}{y}\dfrac{dy}{dx}= \dfrac{a + b}{x + y}\bigg[1 + \dfrac{dy}{dx} \bigg] \\ \end{gathered}
x
a
+
y
b
dx
dy
=
x+y
a+b
[1+
dx
dy
]
\begin{gathered}\rm \: \dfrac{a}{x} + \dfrac{b}{y}\dfrac{dy}{dx}= \dfrac{a + b}{x + y} \: +\dfrac{a + b}{x + y} \: \dfrac{dy}{dx} \\ \end{gathered}
x
a
+
y
b
dx
dy
=
x+y
a+b
+
x+y
a+b
dx
dy
\begin{gathered}\rm \: \dfrac{b}{y}\dfrac{dy}{dx} - \dfrac{a + b}{x + y}\dfrac{dy}{dx}= \dfrac{a + b}{x + y} \: - \: \dfrac{a}{x} \\ \end{gathered}
y
b
dx
dy
−
x+y
a+b
dx
dy
=
x+y
a+b
−
x
a
\begin{gathered}\rm \: \bigg(\dfrac{b}{y} - \dfrac{a + b}{x + y}\bigg)\dfrac{dy}{dx}= \dfrac{a + b}{x + y} \: - \: \dfrac{a}{x} \\ \end{gathered}
(
y
b
−
x+y
a+b
)
dx
dy
=
x+y
a+b
−
x
a
\begin{gathered}\rm \: \bigg(\dfrac{bx + by - ay - by}{y(x + y)} \bigg)\dfrac{dy}{dx}= \dfrac{ax + bx - ax - ay}{x(x + y)} \\ \end{gathered}
(
y(x+y)
bx+by−ay−by
)
dx
dy
=
x(x+y)
ax+bx−ax−ay
\begin{gathered}\rm \: \bigg(\dfrac{bx - ay}{y} \bigg)\dfrac{dy}{dx}= \dfrac{bx - ay}{x} \\ \end{gathered}
(
y
bx−ay
)
dx
dy
=
x
bx−ay
\begin{gathered}\rm \: \bigg(\dfrac{1}{y} \bigg)\dfrac{dy}{dx}= \dfrac{1}{x} \\ \end{gathered}
(
y
1
)
dx
dy
=
x
1
\begin{gathered}\rm \: x\dfrac{dy}{dx} = y \\ \end{gathered}
x
dx
dy
=y
On differentiating w. r. t. x, we get
\begin{gathered}\rm \: \dfrac{d}{dx}\bigg(x\dfrac{dy}{dx}\bigg) = \dfrac{d}{dx}y \\ \end{gathered}
dx
d
(x
dx
dy
)=
dx
d
y
\begin{gathered}\rm \: x\dfrac{d}{dx}\dfrac{dy}{dx} + \dfrac{dy}{dx} \times \dfrac{d}{dx}x = \dfrac{dy}{dx} \\ \end{gathered}
x
dx
d
dx
dy
+
dx
dy
×
dx
d
x=
dx
dy
\begin{gathered}\rm \: x\dfrac{d^{2} y}{d {x}^{2} } + \dfrac{dy}{dx} \times 1 = \dfrac{dy}{dx} \\ \end{gathered}
x
dx
2
d
2
y
+
dx
dy
×1=
dx
dy
\begin{gathered}\rm \: x\dfrac{d^{2} y}{d {x}^{2} } + \dfrac{dy}{dx} = \dfrac{dy}{dx} \\ \end{gathered}
x
dx
2
d
2
y
+
dx
dy
=
dx
dy
\begin{gathered}\rm \: x\dfrac{d^{2} y}{d {x}^{2} } = 0 \\ \end{gathered}
x
dx
2
d
2
y
=0
\begin{gathered}\rm \: \rm\implies \:\dfrac{d^{2} y}{d {x}^{2} } = 0 \\ \end{gathered}
⟹
dx
2
d
2
y
=0
Hence, Proved
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Additional Information
\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}
f(x)
k
sinx
cosx
tanx
cotx
secx
cosecx
x
logx
e
x
dx
d
f(x)
0
cosx
−sinx
sec
2
x
−cosec
2
x
secxtanx
−cosecxcotx
2
x
1
x
1
e
x