Proof By Induction Other Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction:
a) For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
a
Answer and Solution are the same for proofs
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
v
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
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Question by: ada
Answer by: ada
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
Question 1
Prove by mathematical induction:
a) For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
Question 1
Prove by mathematical induction:
a) For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: For any positive integer $n$, $\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^1$
$=1$
$RHS=1\times x^{1-1}$
$=1\times x^0$
$=1\times1$
$=1$
As $LHS=RHS$ it is true when $n=1$
Step 2 - Assume true for $n=k$
So, $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show when $\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}=(k+1)x^{(k+1)-1}$
$LHS=\frac{\mathrm{d}}{\mathrm{d}x}x^{k+1}$
$=\frac{\mathrm{d}}{\mathrm{d}x}{(x\times x^k)}$
$=x\times\frac{\mathrm{d}}{\mathrm{d}x}x^k+x^k\times\frac{\mathrm{d}}{\mathrm{d}x}x$ (by product rule)
$=x\times kx^{k-1}+x^k\times1$ (from Step 2)
$=kx^k+x^k$
$=(k+1)x^k$
$=(k+1)x^{(k+1)-1}$
$=RHS$
Step 4 - Conclusion
It holds that $\frac{\mathrm{d}}{\mathrm{d}x}x^k=(k+1)x^{(k+1)-1}$ when $\frac{\mathrm{d}}{\mathrm{d}x}x^k=kx^{k-1}$. As true when $n=1$ it is also true for all postive integers by mathematical induction.